Question 515 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
Let $\text{x}=\text{a}\cot\theta$
$\Rightarrow\ \text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2\cot^2\theta+\text{a}^2}}\Big\}$
$\Rightarrow\text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2(\cot^2\theta+1)}}\Big\}$
$\Rightarrow\ \text{y}=\sin^{-1}\Big(\frac{\text{a}\cot\theta}{\text{a cosec}\theta}\Big)$
$\Rightarrow\ \text{y}=\cos^{-1}\Bigg(\frac{\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin\theta}}\Bigg)$
$\Rightarrow\ \text{y}=\cos^{-1}(\cos\theta)$
$\Rightarrow\ \text{y}=\theta$
$\Rightarrow\ \text{y}=\cot^{-1}\big(\frac{\text{x}}{\text{a}}\big)\ \big[\text{Since, x}=\text{a}\cot\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{\text{x}}{\text{a}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}^2}{\text{a}^2+\text{x}^2}\times\big(\frac{1}{\text{a}}\big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 525 Marks
If $\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
AnswerHere,
$\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c}$
Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sqrt{\text{y}+\text{x}})+\frac{\text{d}}{\text{dx}}\sqrt{\text{y}-\text{x}}=\frac{\text{d}}{\text{dx}}(\text{c})$
$\Rightarrow\frac{1}{2\sqrt{\text{y}+\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}+\text{x})+\frac{1}{2\sqrt{\text{y}-\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}-\text{z})=0$
$\Rightarrow \frac{1}{2\sqrt{\text{y}+\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}+1\Big)+\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}+\text{x}}}\Big)+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big)=\frac{1}{2\sqrt{\text{y}-\text{x}}}-\frac{1}{2\sqrt{\text{y}+\text{x}}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\times\Big[\frac{1}{\sqrt{\text{y}+\text{x}}}+\frac{1}{\sqrt{\text{y}-\text{x}}}\Big]=\frac{1}{2}\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big[\frac{\sqrt{\text{y}-\text{x}}-\sqrt{\text{y}+\text{x}}}{\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}\Big]=\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}+\sqrt{\text{y}-\text{x}}}\times\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}+\text{x}}}$
[Rationalizing the denominator]
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}+\text{x})+(\text{y}-\text{x})-2\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}{\text{y}+\text{x}-\text{y}+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}-2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{2\text{x}}-\frac{2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2-\text{x}^2}{\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
View full question & answer→Question 535 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
AnswerLet $\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\}$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2+\text{a}^2\sin^2\theta}}\Big\}$
$\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}\Big\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}(1+\cos\theta)}\Big\}$
$=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$=\tan^{-1}\bigg(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\times\frac{\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....\text{(i)}$
Here, $-\text{a}<\text{x}<\text{a}$
$\Rightarrow -1<\frac{\text{x}}{\text{a}}<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}+\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big) \big[\text{Since, x}=\text{a}\sin\theta\big]$
Differentiating it with respect to x using chian rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$=\frac{\text{a}}{2\sqrt{\text{a}^2-\text{x}^2}}\Big(\frac{1}{\text{a}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}$
View full question & answer→Question 545 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
AnswerLet $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^\text{x}$
Also, let $\text{u}=\text{x}^{\sin\text{x}}\text{ and v}=(\sin\text{x})^\text{x}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\sin\text{x}}$
$\Rightarrow\log\text{u}=\log\big(\text{x}^{\sin\text{x}}\big)$
$\Rightarrow\log\text{u}=\sin\text{x}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x})\times\log\text{x}+\sin\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\cot\text{x}\log\text{x}+\sin\text{x}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\Big]\ .....(\text{ii})$
$\text{v}=(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\log(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\text{x}\log(\sin\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log(\sin\text{x})+\text{x}\times\frac{\text{d}}{\text{dx}}\big[\log(\sin\text{x})\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\log(\sin\text{x})+\text{x}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\log\sin\text{x}+\frac{\text{x}}{\sin\text{x}}\cos\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\big)+(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
View full question & answer→Question 555 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big),-1<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big)$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)+\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\sin^{-1}\Big\{\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow\Big(-\frac{\pi}{2}+\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}+\theta\Big)<\frac{3\pi}{4}$
So, from equation (i),
$\text{y}=\theta+\frac{\pi}{4}\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ as }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\sin^{-1}\text{x}+\frac{\pi}{4} \big[\text{Since},\sin\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 565 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy }\log(\text{x}+\text{y})=1$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{xy}\log(\text{x}+\text{y})\big]=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using chain rule and product rule]
$\Rightarrow\text{xy}\times\Big(\frac{1}{\text{x}+\text{y}}\Big)\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
$\Big[\text{Since from equation (i)}\log(\text{x}+\text{y})=\frac{1}{\text{xy}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big)\Big(\frac{(\text{x}+\text{y})\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
$=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
So,
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}^2\text{y}+\text{x}+\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
View full question & answer→Question 575 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{4\text{x}}{1-4\text{x}^2}\Big),-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{4\text{x}}{1-4\text{x}^2}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\{\tan2\theta\}\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow -1<2\text{x}<1$
$\Rightarrow -1<\tan\theta<1$
$\Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow -\frac{\pi}{2}<(2\theta)<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta\Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=2\tan^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\tan\theta\big]$
Differentiating ti with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(2\text{x})^2}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{4}{1+4\text{x}^2}$
View full question & answer→Question 585 Marks
If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
Differentiate with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]+\frac{\text{d}}{\text{dx}}\big[\sin\text{a}\cos(\text{a}+\text{y})\big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\sin\text{a}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})=0$
$\Rightarrow\Big[\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})(1)\Big] \\ +\sin\text{a}\Big[-\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})\Big]=0$
$\Rightarrow\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[-\sin\text{a}\frac{\cos^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}-\sin\text{a}\sin(\text{a}+\text{y})\Big] \\ =-\sin(\text{a}+\text{y})$
$\Big[\because\ \text{x}=-\sin\text{a}\frac{\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big]$
$ \Rightarrow-\frac{\text{dy}}{\text{dx}}\Big[\frac{\sin\text{a}\cos^2(\text{a}+\text{y})+\sin\text{a}\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\Big[\frac{\sin(\text{a}+\text{y})}{\sin\text{a}\{\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})\}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 595 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
Put $\text{x}=\cos2\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\Big(\frac{1}{\sqrt{2}}\Big)\sin\theta\Big\}$
$=\sin^{-1}\Big\{\cos\theta\sin\Big(\frac{\pi}{4}\Big)+\cos\frac{\pi}{4}\sin\theta\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\cos2\theta<1$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0 < \theta < \frac{\pi}{4}$
$\Rightarrow \frac{\pi}{4}<\Big(\theta+\frac{\pi}{4}\Big)<\frac{\pi}{2}$
So from eqaution (i),
$\text{y}=\theta+\frac{\pi}{4}\ \Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}+\frac{\pi}{4}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)+0$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 605 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{2\text{t}}{1+\text{t}^2}\text{ and y}=\frac{1-\text{t}^2}{1+\text{t}^2}$
AnswerHere, $\text{x}=\frac{2\text{t}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dx}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2-2\text{t}^2}{(1+\text{t}^2)}\Big]$
$\frac{\text{dx}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}...(\text{i})$
And, $\text{y}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t})^{2}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{-2\text{t}-2\text{t}^{3}-2\text{t}+2\text{t}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\Big]...(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-4\text{t}}{(1-\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{2(1+\text{t}^{2})}$
$=\frac{-2\text{t}}{(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}} $
$\Big[\text{Since}\frac{\text{x}}{\text{y}}=\frac{2\text{t}}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{1-\text{t}^{2}}=\frac{2\text{t}}{1-\text{t}^{2}}\Big]$
View full question & answer→Question 615 Marks
If $\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere,
$\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
Differentiating it with respect to x using chain rule and quotient rule,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)+\frac{2}{\sqrt{3}}\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}+\frac{2}{\sqrt{3}}\Bigg\{\frac{1}{1+\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)}\Bigg\}\frac{\text{d}}{\text{dx}}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\bigg(\frac{(\text{x}^2-\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg) \\ +\frac{2}{\sqrt{3}}\Big\{\frac{(1-\text{x})^2}{1+\text{x}^4-2\text{x}^2+3\text{x}^2}\Big\}\bigg\{\frac{(1-\text{x}^2)^2\frac{\text{d}}{\text{dx}}(\sqrt{3\text{x}})-\sqrt{3}\text{x}\frac{\text{d}}{\text{dx}}(1-\text{x})^2}{(1-\text{x}^2)^2}\bigg\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{1}{\text{x}^2-\text{x}+1}\Big)\Big(\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{(1-\text{x}^2)^2}{1+\text{x}^2+\text{x}^4}\Big)\Big(\frac{(1-\text{x}^2)(\sqrt{3})-\sqrt{3}\text{x}(-2\text{x})}{(1-\text{x}^2)^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^2-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{\text{x}^4+2\text{x}^2+1-\text{x}^2}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{\sqrt{3}-\sqrt{3}\text{x}^2\sqrt{3}\text{x}^2}{1+\text{x}^2+\text{x}^4}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{-2\text{x}^2+2}{\text{x}^4+\text{x}^2+1}\Big)+\frac{2\sqrt{3}(\text{x}^2+1)}{\sqrt{3}(1+\text{x}^2+\text{x}^4)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2)}{(\text{x}^4+\text{x}^2+1)}+\frac{2(\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2+\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2+\text{x}^4}$
View full question & answer→Question 625 Marks
Differentiate the following functions from first principles:
$\sin^{-1}(2\text{x}+3)$
AnswerLet $\text{f(x)}=\sin^{-1}(2\text{x}+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2(\text{x}+\text{h})+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2\text{x}+2\text{h}+3)$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}(2\text{x}+2\text{h}+3)-\sin^{-1}(2\text{x}+3)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\Big[(2\text{x}+2\text{h}+3)\sqrt{1+(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big]}{\text{h}}$
$\Big[\text{Since}, \sin^{-1}\text{x}-\sin^{-1}\text{y}=\sin^{-1}\big[\text{x}\sqrt{1-\text{y}^2}-\text{y}\sqrt{1-\text{x}^2}\big]\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{z}}{\text{z}}\times\frac{\text{z}}{\text{h}}$
Where, $\text{z}=(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}$
$\text{and }\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{h}}{\text{h}}=1$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{z}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)^2-(2\text{x}+3)^2-(2\text{x}+3)^2\big(1-(2\text{x}+2\text{h}+3)^2\big)}{\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
[Since, rationalizing numerator]
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)\big]\big(1-(2\text{x}+3)^2\big)-(2\text{x}+3)^2\big[1-(2\text{x}+3)^2-4\text{h}^2-4\text{h}(2\text{x}+3)\big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)-(2\text{x}+3)^4-4\text{h}^2(2\text{x}+3)^2-4\text{h}(2\text{x}+3)^3 -(2\text{x}+3)^2+(2\text{x}+3)^3+4\text{h}^2(2\text{x}+3)^2+4\text{h}(2\text{x}+3)^3\Big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{\text{h}\Big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big\}}$
$=\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{4(2\text{x}+3)}{2(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\sin^{-1}(2\text{x}+3)\big)=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
View full question & answer→Question 635 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$ with respect to $\sin^{-1}\big(3\text{x}-4\text{x}^3\big),$ if $-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$Put $\text{x}=\tan\theta$
$\Rightarrow \text{u}=\tan^{-1}\Big(\frac{\tan\theta-1}{\tan\theta+1}\Big)$
$\Rightarrow \text{u}=\tan^{-1}\bigg(\frac{\tan\theta-\tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}\bigg)$
$\Rightarrow\text{u}\tan^{-1}\Big[\tan\big(\theta-\frac{\pi}{4}\big)\Big]\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}2{}<\tan\theta<\frac{1}{2}$
$\Rightarrow-\tan^{-1}\big(\frac{1}{2}\big)<\theta<\tan^{-1}\big(\frac{1}{2}\big)$
So, from equation (i)
$\text{u}=\theta-\frac{\pi}{4}$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\tan^{-1}\text{x}-\frac{\pi}{4} \ [\text{Since,x}= \tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}-0$
$\Rightarrow \frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ ..... \text{(ii)}$
And,
Let, $\text{v}=\sin^{-1}(3\text{x}-4\text{x}^3)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{v}=\sin^{-1}(3\sin\theta-4\sin^3\theta)$
$\Rightarrow\text{v}=\sin^{-1}(\sin3\theta)\ .....\text{(iii)}$
Now, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}{2}<\sin\theta<\frac{1}{2}$
$\Rightarrow-\frac{1}{6}<\theta<\frac{\pi}{6}$
So, from equation (iii),
$\text{v}=3\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=3\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Dirrerentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{3}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x}^2}}{3}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{3(1+\text{x}^2)}$
View full question & answer→Question 645 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2\sqrt{2}}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(\frac{\pi}{4},\frac{3\pi}{4}\Big)$
So, from equation (i),
$\text{u}=\pi=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big(\frac{\pi}{2},\pi\Big)\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}\ .....(\text{iii})$
From equation (ii)
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{iv})$
Diferentiating equation (ii) with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-4\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}(-8\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (iii) by (v)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 655 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
AnswerLet $\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
Also, let $\text{u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \text{y}=\text{v}+\text{u}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}+\frac{\text{du}}{\text{dx}}\ .....(\text{i})$
Now, $\text{u}=(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\log\big[(\log\text{x})^\text{x}\big]$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big[\log(\log\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log(\log\text{x})+\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{\text{x}}{\log\text{x}}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]\ .....(\text{ii})$
Also, $\text{v}=\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}\log\text{x}=(\log\text{x})^2$
Differentiating both sides with respect to x,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\log)^2\big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{v}(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}+(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]$
View full question & answer→Question 665 Marks
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}$ and $\text{y}=\frac{\cos^{3}\text{t}}{\sqrt{\cos2\text{t}}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}\Big] $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\sin^{3})-\sin^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\sin^{3}\text{t})\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^{3}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}(\sin^{2}\text{t}\cos\text{t})-\frac{\sin^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\cos2\text{t}\sin^{\text{2}}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
Now, $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^{3}}{\sqrt{\cos2\text{t}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}} \frac{\text{d}}{\text{dt}}(\cos^{3}\text{t})-\cos^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\cos^{2}\text{t})\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^{3}\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}\cos^{2}\text{t}-(\sin\text{t})-\frac{\cos^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}\sin\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}\times\frac{\cos2\text{t}\sqrt{\cos2\text{t}}}{3\cos2\text{t}\sin^{2}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sin\text{t}\cos{\text{t}}[-3\cos2\text{t}\cos\text{t}+2\cos^{3}\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin{\text{t}}+2\sin^{3}\text{t}]}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{[-3(2\cos^{2}\text{t}-1)\cos\text{t}+2\cos^{3}\text{t}]}{[3(1-2\sin^{2}\text{t})\sin\text{t}+2\sin^{3}\text{t}]}$
$\begin{bmatrix} \cos2\text{t}=2\cos^2\text{t}-1 \\ \cos2\text{t}=1-2\sin^2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-4\cos^{3}\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^{3}\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-\cos3\text{t}}{\sin3\text{t}}$
$\begin{bmatrix} \cos3\text{t}=4\cos^3\text{t}-3\cos\text{t} \\ \sin3\text{t}=3\sin^2\text{t}-4\sin^3\text{t} \end{bmatrix}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\cos3\text{t}$
View full question & answer→Question 675 Marks
Differentiate the following functions with respect to x:
$\sin^2\{\log(2\text{x}+3)\}$
AnswerConsider $\text{y}=\sin^2\{\log(2\text{x}+3)\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2\{\log(2\text{x}+3)\}\big]$
$=2\sin\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}\sin\big(\log(2\text{x}+3)\big)$
[Using chain rule]
$=2\sin\big(\log(2\text{x}+3)\big)\cos\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{1}{(2\text{x}+3)}\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$\big[\text{Since}, 2\sin\text{A}\cos\text{A}=\sin^2\text{A}\big]$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2\log(2\text{x}+3)\big)=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
View full question & answer→Question 685 Marks
Differentiate the following functions with respect to x:
$\sin(\text{x}^\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Taking log on both sides,
$\log(\sin^{-1}\text{y})=\log\text{x}^\text{x}$
$\Rightarrow\ \log(\sin^{-1}\text{y})=\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\frac{\text{dy}}{\text{dy}}(\sin^{-1}\text{y})=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\times\Big(\frac{1}{\sqrt{1-\text{y}^2}}\Big)\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{y}\sqrt{1-\text{y}^2}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}(\sin\text{x}^\text{x})\sqrt{1-(\sin\text{x}^\text{x})^2}(1+\log\text{x})$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cos\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
View full question & answer→Question 695 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
AnswerHere,
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
$=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{x}^\frac{1}{\text{x}}}$
$\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\big(\frac{1}{\text{x}}\log\text{x}\big)}$
$\big[\text{Since, e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\frac{1}{\text{x}}\log\text{x}}\Big)$
$=\text{e}^{\text{x}\log\text{x}}+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\frac{1}{\text{x}}\log\text{x}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\log\text{x}\Big)$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{x}^\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +\text{x}^\frac{1}{\text{x}}\Big[\Big(\frac{1}{\text{x}}\Big)\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\Big(\frac{1}{\text{x}^2}-\frac{1}{\text{x}^2}\log\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\frac{(1-\log\text{x})}{\text{x}^2}$
View full question & answer→Question 705 Marks
If $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big),\text{x}>0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Rightarrow \text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)+\cos^{-1}\Big(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\Big)$
$\Rightarrow \text{y}=\tan^{-1}(\tan2\theta)+\cos^{-1}(\cos2\theta)$
$\Rightarrow \text{y}=2\theta+2\theta$
$\Rightarrow \text{y}=4\theta$
$\Rightarrow \text{y}=4\tan^{-1} \text{x} \big[\text{using, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
View full question & answer→Question 715 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
AnswerWe have, $\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}+\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\cos\text{x}\log\sin\text{x}}+\text{e}^{\sin\text{x}\log\cos\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}\log\cos\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}\big(\cos\text{x}\log\sin\text{x})+\text{e}^{\sin\text{x}\log\cos\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\cos\text{x})$
$=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}\Big[\sin\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\Big[\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\times(-\sin\text{x})\Big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\sin\text{x}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\times(\cos\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\tan\text{x}(-\sin\text{x})+\cos\text{x}\log\cos\text{x}\big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}\big]$
View full question & answer→Question 725 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}-\text{a}}{\text{x}}}{\frac{\text{x}+\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}}{\text{x}}-\frac{\text{x}}{\text{x}}}{\frac{\text{x}}{\text{x}}+\frac{\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{1-\frac{\text{x}}{\text{x}}}{1+1\times\frac{\text{a}}{\text{x}}}\bigg)$
$\text{y}=\tan^{-1}(1)-\tan^{-1}\big(\frac{\text{a}}{\text{x}}\big)$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{1+\big(\frac{\text{a}}{\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{a}}{\text{x}}\big)$
$=-\frac{\text{x}^2}{\text{x}^2+\text{a}^2}\Big(\frac{-\text{a}}{\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 735 Marks
Differentiate the following functions with respect to x:
$(\text{x}^\text{x})\sqrt{\text{x}}$
AnswerLet $\text{y}=(\text{x}^\text{x})\sqrt{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\text{x}^\text{x}\sqrt{\text{x}})$
$\log\text{y}=\text{x}\log\text{x}+\frac{1}{2}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)+\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}+\frac{1}{2\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\sqrt{\text{x}}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}+\frac{1}{2}}\Big[\Big(\frac{2\text{x}+1}{2\text{x}}\Big)+\log\text{x}\Big]$
View full question & answer→Question 745 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big), \frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\tan^{-1}\Big[\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big]$
$=\tan\bigg[\frac{\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\bigg[\frac{\frac{\cos\text{x}}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\Big[\frac{1+\tan\text{x}}{1-\tan\text{x}}\Big]$
$=\tan^{-1}\bigg[\frac{\frac{\tan\pi}{4}+\tan\text{x}}{1-\frac{\tan\pi}{4}\tan\text{x}}\bigg]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\text{x}\Big)\Big]$
$\text{y}=\frac{\pi}{4}+\text{x}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 755 Marks
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
Answerconsider the given function,
$\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1$
Differentiating both sides w.r.t. 'x' we get
$-\sin\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(-\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}\Big)+\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos(\text{a}+\text{y})}{\text{x}\sin(\text{a}+\text{y})-\sin\text{y}}$
Multiplying the numerator and the denominator
by $\cos(\text{a}+\text{y})$ on the R.H.S., we have,
$\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$\big[\because\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{given function}\big]$
$=\frac{\cos^2(\text{a}+\text{y})}{\sin\big[(\text{a}+\text{y})-\text{y}\big]}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 765 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
AnswerLet, $\text{y}=\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\Big]$
$=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)$
[Using chain rule and quotient rule]
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{\text{x}^2-\text{x}+1\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^3-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+\text{x}+1)(\text{x}^2+\text{x}+1)}$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+1)^2-(\text{x})^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+1+2\text{x}^2-\text{x}^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\Big\}=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
View full question & answer→Question 775 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\},-\frac{3\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\bigg\{\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\bigg\}$
$=\sin^{-1}\Big\{\sin{\text{x}}\cos\frac{\pi}{4}+\cos\text{x}\times\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\text{x}+\frac{\pi}{4}\Big)\Big\}$
Here, $\frac{-3\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\Big(\frac{-3\pi}{4}+\frac{\pi}{4}\Big)$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta, \text{ if }\theta\in\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=1+0$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 785 Marks
Find the derivative of the function $f(x)$ given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f(1).$
AnswerHere,
$f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Differentiating with respect to x using product rule and chain rule,
$\Rightarrow\text{f}'\text{(x)}=(1+\text{x})\big(1+\text{x}^2\big) \frac{\text{d}}{\text{dx}}\big(1+\text{x}^8\big)+(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^8\big) \\ \frac{\text{d}}{\text{dx}}\big(1+\text{x}^4\big)+(1+\text{x})\big(1+\text{x}^4\big)\big(1+\text{x}^8\big) \\ \frac{\text{d}}{\text{dx}}\big(1+\text{x}^2\big)+\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)\big(1+\text{x}^8\big) \frac{\text{d}}{\text{dx}}(1+\text{x})$
$\Rightarrow\text{f}'\text{(x)}=(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)8\text{x}^7+(1+\text{x})\big(1+\text{x}^2\big)\big(1+\text{x}^2)\big(1+\text{x}^8\big)\big(4\text{x}^3\big) \\ +(1+\text{x})\big(1+\text{x}^4\big)\big(1+\text{x}^8\big)(2\text{x})+\big(1+\text{x}^2\big)\big(1+\text{x}^4\big)\big(1+\text{x}^8\big)(1)$
$f'(1)=(1+1)(1+1)(8) + (1+1)(1+1)(1+1)(4) + (1+1)(1+1)(1+1)(2) + (1+1)(1+1)(1+1)$
$f'(1) = (2)(2)(2)(8) + (2)(2)(2)(4) + (2)(2)(2)(2) + (2)(2)(2)$
$= 64 + 32 + 16 + 8$
$=120$
So,
$f'(1) = 120$
View full question & answer→Question 795 Marks
If $x^{16}y^9 = (x + y)^{17}$, prove that $\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
AnswerHere,$x^{16}y^9 = (x + y)^{20}$
Taking log on both the siede,
$\log(\text{x}^{16}\times\text{y}^{19})=\log(\text{x}^2+\text{y})^{17}$
$\big[\text{since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$16\log\text{x}+9\log\text{y}=17\log(\text{x}^2+\text{y})$
Differentiating it with respect to x using chain rule
$16\frac{\text{d}}{\text{dx}}(\log\text{x})+9\frac{\text{d}}{\text{dx}}(\log\text{y})=17\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=17\frac{1}{(\text{x}^2+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{17}{\text{x}^2+\text{y}}\Big[2\text{x}+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{17}{(\text{x}^2+\text{y})}\frac{\text{dy}}{\text{dx}}=\Big(\frac{34\text{x}}{\text{x}^2+\text{y}}\Big)-\frac{16}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9}{\text{y}}-\frac{17}{(\text{x}^2+\text{y})}\Big]=\frac{34\text{x}^2-16\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9\text{x}^2+9\text{y}-17\text{y}}{\text{y}(\text{x}^2+\text{y})}\Big]=\frac{18\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\frac{2(9\text{x}^2-8\text{y})}{9\text{x}^2-8\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
View full question & answer→Question 805 Marks
If $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=10(1-\cos\text{t})\ ...(\text{i})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=12(\sin\text{t})\ ...(\text{ii})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{12(\sin\text{t})}{10(1-\cos\text{t})}$ From equation (i) and (ii)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{12\sin\frac{\text{t}}{2}\cdot\cos\frac{\text{t}}{2}}{10\sin^2\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
View full question & answer→Question 815 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$
AnswerHere,
$\text{y}=(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$=\text{e}^{\log(\sin\text{x})^\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\text{y}=\text{e}^{\text{x}\log\sin\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\big[\text{Since},\log_\text{e}^\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentitating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\sin^{-1}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})+\frac{1}{\sqrt{1-\big(\sqrt{\text{x}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\Big]$
$=(\sin\text{x})^\text{x}\Big[\text{x}\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(1)\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$=(\sin\text{x})^\text{x}\Big[\frac{\text{x}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\text{x}\cot\text{x}+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
View full question & answer→Question 825 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
AnswerLet, $\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$=\frac{1}{\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\Big(\frac{1+\cos\text{x}}{\sin\text{x}}\Big)\Bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})-\sin\text{x}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\Bigg]$
[Using quotient rule]
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\bigg[\frac{(1+\cos\text{x})(\cos\text{x})-\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{(1+\cos\text{x})}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\Big)=\text{cosec x}$
View full question & answer→Question 835 Marks
Differentiate the following functions from first principles:
$\log\text{cosec x}$
AnswerLet $\text{f(x)}=\log\text{cosec x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\text{cosec x}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\text{cosec}(\text{x}+\text{h})-\log\text{cosec x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{\frac{\text{cosec}(\text{x}+\text{h})}{\text{cosec x}}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}}{\sin(\text{x}+\text{h})}-1\Big)\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\begin{Bmatrix}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big)\Big\}}{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}} \end{Bmatrix}\frac{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{\text{x}+\text{x}+\text{h}}{2}\Big)\sin\Big(\frac{\text{x}-\text{x}-\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})\text{h}}$
$\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\text{ and }\sin\text{A}-\sin\text{B} \\ =2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})(-2)}\bigg\{\frac{\sin\big(-\frac{\text{h}}{2}\big)}{-\frac{\text{h}}{2}}\bigg\}$
$=-\cot\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}(\log\text{cosec x})=-\cot\text{x}$
View full question & answer→Question 845 Marks
If $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$ prove that $\frac{\text{dy}}{\text{dx}}=\log\Big(\frac{\text{x}-1}{1+\text{x}}\Big)$
AnswerWe have, $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)\big]$
$=\Big[(\text{x}-1)\frac{\text{d}}{\text{dx}}\log(\text{x}-1)+\log(\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}-1)\Big] \\ -\Big[(\text{x}+1)\frac{\text{d}}{\text{dx}}\log(\text{x}+1)+\log(\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}+1)\Big]$
$=\Big[(\text{x}-1)\times\frac{1}{(\text{x}-1)}\frac{\text{d}}{\text{dx}}(\text{x}-1)+\log(\text{x}-1)\times(1)\Big] \\ -\Big[(\text{x}+1)\times\frac{1}{(\text{x}+1)}\times\frac{\text{d}}{\text{dx}}(\text{x}+1)+\log(\text{x}+1)(1)\Big]$
$=\big[1+\log(\text{x}-1)\big]-\big[1+\log(\text{x}+1)\big]$
$=\log(\text{x}-1)-\log(\text{x}+1)$
$=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
View full question & answer→Question 855 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with resepect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log\text{x}.\log(\log\text{x})\big]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x}).\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x}).\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x})+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$
View full question & answer→Question 865 Marks
If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
AnswerWe have, $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\sin\text{x})^\text{y}=\log(\cos\text{y})^{\text{x}}$
$\Rightarrow\text{y}\log(\sin\text{x})=\text{x}\log(\cos\text{y})$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\big[\text{y}\log\sin\text{x}\big]=\frac{\text{d}}{\text{dx}}\big[\text{x}\log\cos\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\text{x}\frac{\text{dy}}{\text{dx}}(\log\cos\text{y})+\log\cos\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\text{y}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}\frac{\text{x}}{\text{dx}}(\cos\text{y})+\log\cos\text{y}(1)$
$\Rightarrow\frac{\text{y}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}(-\sin\text{y})\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow \text{y}\cot\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\sin\text{x}+\text{x}\tan\text{y}) \\ =\log\cos\text{y}-\text{y}\cot\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 875 Marks
Differentiate the following functions with respect to x:
$(\sin^{-1}\text{x})^\text{x}$
AnswerLet $\text{y}=(\sin^{-1}\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin^{-1}\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(\sin^{-1}\text{x})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\sin^{-1}\text{x})+\log\sin(-1)\times\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\times\frac{1}{\sin^{-1}\text{x}}\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\log\sin^{-1}\text{x}(1)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sin^{-1}\text{x}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\log\sin^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\bigg[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\big(\sqrt{1-\text{x}^2}\big)}\bigg]$
$\frac{\text{dy}}{\text{dx}}=(\sin^{-1}\text{x})^2\Big[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 885 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2-1}}{\text{ax}}\Big),\text{x}\neq0$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
Put $\text{ax}=\tan\theta$
$\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$
$=\frac{\theta}{2}$
$\text{y}=\frac{1}{2}\tan^{-1}(\text{ax})$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\Big(\frac{1}{1+(\text{ax})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{ax})$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2(1+\text{a}^2\text{x}^2)}(\text{a})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{2(1+\text{a}^2\text{x}^2)}$
View full question & answer→Question 895 Marks
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
AnswerLet $\text{y}=\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
$\Rightarrow\text{y}=\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\big[\log(\text{x}-1)-\log(\text{x}+1)\big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}-1)\big\}-\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}+1)\big\}\Big]$
$=\frac{1}{2}\Big(\frac{1}{\text{x}-1}-\frac{1}{\text{x}+1}\Big)$
$=\frac{1}{2}\Big(\frac{2}{\text{x}^2-1}\Big)$
$=\frac{2}{\text{x}^2-1}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}^2-1}$
View full question & answer→Question 905 Marks
Differentiate the following functions from first principles:
$x^2e^x$.
AnswerLet $f(x) = x^2e^x$
$\Rightarrow f(x + h) = (x + h)^2 e^{(x+h)}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{x}^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^\text{x}}{\text{h}}+\frac{2\text{xhe}^{(\text{x}+\text{h})}}{\text{h}}+\frac{\text{h}^2\text{e}^{(\text{x}+\text{h})}}{\text{h}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}^2\text{e}^\text{x}\big(\text{e}^{(\text{x}+\text{h}-\text{x})}-1\big)}{\text{x}}+2\text{xe}^{(\text{x}+\text{h})}+\text{he}^{(\text{x}+\text{h})}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{x}^2\text{e}^{\text{x}}\frac{\big(\text{e}^\text{h}-1\big)}{\text{h}}+2\text{xe}^{(\text{x}+\text{h})}+\text{h}^{\text{e}}(\text{x}+\text{h})\bigg]$
$=\text{x}^2\text{e}^\text{x}+2\text{xe}^\text{x}+0\times\text{e}^\text{x}\ \Big[\text{Since,}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x})=\text{e}^\text{x}(\text{x}^2+2\text{x})$
View full question & answer→Question 915 Marks
If $\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}$
AnswerHere,
$\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\log(\text{x}^2+\text{y}^2)^{\frac{1}{2}}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\frac{1}{2}\log(\text{x}^2+\text{y}^2)=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
Differentiating with respect to x,
$\Rightarrow\frac{1}{2}\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y}^2)=\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\times\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\frac{1}{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$ \Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\times2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\bigg]$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{y}-\text{x})=-(\text{y}+\text{x})$
View full question & answer→Question 925 Marks
Differentiate the following functions with respect to x:
$(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\cos\text{x})^\text{x}}+\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log(\cos\text{x})}+\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\cos\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}\big)$
$=\text{e}^{\log\cos\text{x}}\times\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{x})+\text{e}^{\frac{1}{\text{x}}\log\sin}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\log\sin\text{x}\big)$
$=\text{e}^{\log(\cos\text{x})^\text{x}}\times\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\times\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}\times\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}+\log\cos\text{x}(1)\Big] \\ +(\sin)^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\cos\text{x}}\Big)(-\sin\text{x})+\log\cos\text{x}\Big] \\ +(\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}(\cos\text{x})-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
$=(\cos\text{x})^\text{x}\big[\log\cos\text{x}-\text{x}\tan\text{x}\big](\sin\text{x})^\frac{1}{\text{x}} \\ \Big[\frac{\cot\text{x}}{\text{x}}-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
View full question & answer→Question 935 Marks
Differentiate the following functions with respect to x:
$\text{e}^\text{x}\log\sin2\text{x}$
AnswerLet $\text{y}=\text{e}^\text{x}\log\sin2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{e}^\text{x}\log\sin2\text{x}\big]$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\log\sin2\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using product rule and chain rule]
$=\text{e}^\text{x}\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\log\sin2\text{x}\big(\text{e}^\text{x}\big)$
$=\frac{\text{e}^\text{x}}{\sin2\text{x}}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^\text{x}\log\sin2\text{x}$
$=\frac{2\cos2\text{xe}^\text{x}}{\sin2\text{x}}+\text{e}^\text{x}\log\sin2\text{x}$
$=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\log\sin2\text{x}\big)=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
View full question & answer→Question 945 Marks
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
AnswerWe have, $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{y}\sqrt{\text{x}^2+1}\Big)=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$
[Using Product rule and chain rule]
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$
$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$
$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$
View full question & answer→Question 955 Marks
If $y^x = e^{y-x}$, Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
AnswerHere,
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
View full question & answer→Question 965 Marks
Differentiate the following functions with respect to x:
$\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
AnswerLet $\text{y}=\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big]$
$\frac{\text{d}}{\text{dx}}\log(3\text{x}+2)-\frac{\text{d}}{\text{dx}}\big(\text{x}^2\log(2\text{x}-1)\big)$
$=\frac{1}{3\text{x}+2}\frac{\text{d}}{\text{dx}}(3\text{x}+2)-\Big[\text{x}^2\frac{\text{d}}{\text{dx}}\log(2\text{x}-1)+\log(2\text{x}-1)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)\Big]$
[Using product rule and chain rule]
$=\frac{3}{3\text{x}+2}\Big[\text{x}^2\times\frac{1}{2\text{x}-1}\frac{\text{d}}{\text{dx}}(2\text{x}-1)+\log(2\text{x}-1)\times2\text{x}\Big]$
$=\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
So,
$\frac{\text{d}}{\text{dx}}\big(\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big) \\ =\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
View full question & answer→Question 975 Marks
If $x^x + y^x = 1$, prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{y}+\text{x}\log\text{y})}{\text{x}(\text{y}\log\text{x}+\text{x})}$
AnswerHere,
$x^x + y^x = 1$
Taking on bith sides,
$\log(\text{x}^\text{y}\times\text{y}^\text{x})=\log(1)$
$\text{y}=\log\text{x}+\text{x}\log\text{y}=\log1$
$\big[\text{Since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\log1)$
$\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}\text{(x)}\Big]=0$
$\Big[\text{y}\Big(\frac{1}{\text{x}}\Big)\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\Big(\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)+\log\text{y}(1)\Big]=0$
$\frac{\text{y}}{\text{x}}+\log\text{x}\log\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}=0$
$\frac{\text{dy}}{\text{dx}}\Big(\log\text{x}+\frac{\text{x}}{\text{y}}\Big)=-\Big[\log\text{y}+\frac{\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{y}\log\text{x}+\text{x}}{\text{y}}\Big]=-\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{y}\log\text{x}+\text{x}}\Big]$
View full question & answer→Question 985 Marks
If $\text{x}=\text{a}\sin2\text{t}(1+\cos 2\text{t})$ and $\text{y}=\text{b}\cos\text{t}(1-\cos2\text{t}),$ show that at $\text{t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}\text{ t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}$
AnswerConsider the given functions,
$\text{x}=\text{a}\sin(2\text{t})(1+\cos2\text{t})\text{ and y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
Rewriting the above function, we have,
$\text{x}=\text{a}\sin2\text{t}+\frac{\text{a}}{2}\sin4\text{t}$
Differentiating the above function w.r.t. 't', we have,
$\frac{\text{dx}}{\text{dx}}=2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}\ .....(\text{i})$
$\text{y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
$\text{y}=\text{b}\cos2\text{t}-\text{b}\cos^22\text{t}$
$\frac{\text{dy}}{\text{dt}}=-2\text{b}\sin2\text{t}+2\text{b}\cos2\text{t}\sin2\text{t} \\ =-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}\ .....(\text{ii})$
From (1) and (2),
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}}{2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}}$
$\therefore\frac{\text{dy}}{\text{dx}}\Big|_{\frac{\pi}{4}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}\Bigg|_{\text{t}=\frac{\pi}{4}}=\frac{-2\text{b}}{-2\text{a}}=\frac{\text{b}}{\text{a}}$
View full question & answer→Question 995 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiate with respect to y,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\{\sin(\text{a}+\text{y})\}+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule and chain rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}+\sin(\text{a}+\text{y})(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\{1-\text{x}\cos(\text{a}+\text{y})\}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\text{x}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y}){-\text{y}\cos(\text{a}+\text{y})}}$
Hence, proved.
View full question & answer→Question 1005 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = e^x + 10^x + x^x$
AnswerHere, $\text{y}=\text{e}^\text{x}+10^\text{x}+\text{x}^\text{x}$
$\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
$\big[\text{Since, e}^{{\log}_\text{a}^\text{b}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
Differentiating it with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(10^{\text{x}})+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[1+\log\text{x}]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[\log\text{e}+\log\text{x}] \big[\text{Since}, \log_\text{e}\text{e}=1\big]$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^\text{x}(\log\text{ex})\ \big[\text{Since}\log\text{A}+\log\text{B}=\log\text{AB}]$
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