Question 2015 Marks
If $e^x + x^y = e^{x+y},$ prove that $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
AnswerHere,
$e^x + e^y = e^{x+y} ......(i)$
Differentiating both the sides using chain rule,
$\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^\text{y})=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}+\text{y}})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{d}}{\text{dx}}\Big]$
$\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\Big(\frac{\text{e}^\text{x}+\text{e}^\text{x}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}-\text{e}^\text{y}}\Big)$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{y}-\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
View full question & answer→Question 2025 Marks
Differentiate $(\log\text{x})^\text{x}$ with respect to x.
AnswerLet $\text{u}=(\log1+\text{x})^\text{x}$
Taking log on both sides,
$\log\text{u}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\big\{\log(\log\text{x})\big\}+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\Big(\frac{1}{\log\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\text{x}}{\log\text{x}}\big(\frac{1}{\text{x}}\big)+\log\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]\ .....(\text{i})$
Again, let $\text{v}=\log\text{x}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}}\ .....(\text{ii})$
Dividing equation (i) by (ii), we get
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\frac{(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}(\log\log\text{x})}{\log\text{x}}\Big]}{\frac{1}{\text{x}}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\text{x}(\log\text{x})^{\text{x}{-1}}(1+\log\text{x}\times\log\log\text{x})$
View full question & answer→Question 2035 Marks
If $\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
Answer$\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}}$
Taking logarithm on both sides, we get
$\log\text{x}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{y}\log\text{x}=\text{x}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=1$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{y}}{\text{x}}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}}$
$\Rightarrow \text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
View full question & answer→Question 2045 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And,
Let $\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\sec^{-1}\Big(\frac{1}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\sec^{-1}(\sec\theta)$
$\Rightarrow\text{v}=\cos^{-1}\bigg(\frac{1}{\frac{1}{\cos\theta}}\bigg)\ \Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}(\cos\theta)\ .....(\text{ii})$
Here,
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\sin\theta\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
Let, $\text{u}=2\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, from equation (ii),
$\text{v}=\theta\big[\text{Since},\cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}\big[\text{Since},\text{x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 2055 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(0, 1)$
AnswerLet $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Now, $\text{x}\in(0,1)$
$\Rightarrow\cos\theta\in(0,1)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{2}\Big)$
So, from equation (i),
$\text{u}=\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\cos^{-1}\text{x}\big[\text{Since},\cos\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\cos^{-1}\text{x}$
Differentaiting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\therefore\frac{\text{du}}{\text{dx}}=1$
View full question & answer→Question 2065 Marks
Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\big(\frac{1}{2}\big)$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{2}\ .....(\text{i})$
Let, $\text{v}=\sec^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}\sqrt{\text{x}^2-1}}\ .....\text{(ii)}$
Differentiating equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{2}\times\frac{\text{x}\sqrt{\text{x}^2-1}}{1}$
$\frac{\text{du}}{\text{dv}}=\frac{-\text{x}\sqrt{\text{x}^2-1}}{2}$
View full question & answer→Question 2075 Marks
If $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{e}^{\text{x}+\text{y}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^{\text{x}}(\text{e}^\text{y}-1)}{\text{e}^{\text{y}}(\text{e}^{\text{x}}-1)}$ or $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
Answer$\text{e}^\text{x}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow\text{e}^\text{x}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}(\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}})=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}}{\text{x}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}({1-\text{e}}^\text{x})}$
$=-\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}(\text{e}^\text{x}-1)}$
View full question & answer→Question 2085 Marks
If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ prove that $2\text{x}\frac{\text{dy}}{\text{dx}}=\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}$
Answer$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{x}^{-1\frac{1}{2}}\Big)$
$=\frac{1}{2\sqrt{\text{x}}}+\Big(-\frac{1}{2}\times\text{x}^{-\frac{1}{2}-1}\Big)$
$=\frac{2}{2\sqrt{\text{x}}}-\frac{1}{2\sqrt[\text{x}]{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}\sqrt{\text{x}}}$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
Hence, the solution is, $2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
View full question & answer→Question 2095 Marks
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ with $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have, $\cos\text{y}=\text{x}\cos(\text{a}+\text{y})$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\cos\text{y})=\frac{\text{d}}{\text{dx}}\big\{\text{x}\cos(\text{a}+\text{y})\big\}$
$\Rightarrow -\sin\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})$
$\Rightarrow -\sin\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})+\text{x}\big[-\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\sin(\text{a}+\text{y})-\sin\text{y}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y}) \\ \Big[\because \cos\text{y}=\text{x}\cos(\text{a}+\text{y})\Rightarrow\text{x}=\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\Big] $
$\Rightarrow\big[\cos\text{y}\sin(\text{a}+\text{y})-\sin\text{y}\cos(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \sin(\text{a}+\text{y}-\text{y})\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 2105 Marks
If $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}},$ prove that $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}$
AnswerGivne, $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\bigg[\frac{\sqrt{1-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x}\sin^{-1}\text{x})-(\text{x}\sin^{-1}\text{x})\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})}{(\sqrt{1-\text{x}^2})^2}\bigg]$
[Using quotient rule, product rule, chain rule]
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}-\big(\text{x}\sin^{-1}\text{x}\big)\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(1-\text{x}^2\big)}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big\}-\frac{\text{x}\sin{-1}\text{x}(-2\text{x})}{2\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\text{x}+\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\sqrt{1-\text{x}^2}\sin^{-1}}{1}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{(1-\text{x}^2)\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}-\text{x}^2\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}\ \Big\{\text{Since, given y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big\}$
View full question & answer→Question 2115 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
AnswerLet $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)-(\text{x}^2+2)\frac{\text{d}}{\text{dx}}(\sqrt{\cos\text{x}})}{(\sqrt{\cos\text{x}})^2}$
[Using quotient rule and chain rule]
$=\frac{2\text{x}\sqrt{\cos\text{x}}-(\text{x}^2+2)\Big(-\frac{1}{2}\frac{\sin\text{x}}{\sqrt{\cos\text{x}}}\Big)}{\cos\text{x}}$
$=\frac{2\text{x}\sqrt{\cos\text{x}}+\frac{(\text{x}^2+2)\sin\text{x}}{2\sqrt{\cos\text{x}}}}{\cos\text{x}}$
$=\frac{4\text{x}\cos\text{x}+(\text{x}^2+2)\sin\text{x}}{2(\cos\text{x})^\frac{3}{2}}$
$=\frac{2\text{x}}{\sqrt{\cos\text{x}}}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{(\cos\text{x})^\frac{3}{2}}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{\cos\text{x}}\Big\}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\tan\text{x}}{2}\Big\}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}\Big)=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\sin\text{x}}{2}\Big\}$
View full question & answer→Question 2125 Marks
Prove that $\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
Answer$\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
$\text{L.H.S}=\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\text{x}\frac{\text{d}}{\text{dx}}\sqrt{\text{a}^2-\text{x}^2}+\sqrt{\text{a}^2-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\frac{\text{a}^2}{2}\times\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\text{x}}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
[Using product rule, chain rule]
$=\frac{1}{2}\bigg[\text{x}\times\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)+\sqrt{\text{a}^2-\text{x}^2}\Big] \\ +\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2}}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\frac{\text{x}(-2\text{x})}{2\sqrt{\text{a}^2-\text{x}^2}}+\sqrt{\text{a}^2-\text{x}^2}\Big]+\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\bigg[\frac{-2\text{x}^2+2\big(\text{a}^2-\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{1}{2}\bigg[\frac{2\big(\text{a}^2-2\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2+\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\big(\text{a}^2-\text{a}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{(\text{a}^2-\text{x}^2)}{\sqrt{\text{a}^2-\text{x}^2}}$
$=\sqrt{\text{a}^2-\text{x}^2}$
$=\text{R.H.S}$
View full question & answer→Question 2135 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
AnswerWe have, $\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
Differentiating it with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{2}{3}}\Big)+\frac{\text{d}}{\text{dx}}\Big(\text{y}^{\frac{2}{3}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\text{a}^{\frac{2}{3}}\Big)$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{2}{3}-1}+\frac{2}{3}\big(\text{y}\big)^{\frac{2}{3}-1}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}+\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}\times\frac{3}{2\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{-1}{3}}}{\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{1}{3}}}{\text{y}^{\frac{1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\big(\frac{\text{x}}{\text{y}}\big)^{\frac{1}{3}}$
View full question & answer→Question 2145 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\big\{2\text{x}\sqrt{1-\text{x}^2}\big\},\frac{1}{\sqrt{2}}<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{2\text{x}\sqrt{1-\text{x}^2}\Big\}$
Put $\text{x}=\cos\theta$
$\text{y}=\cos^{-1}\Big\{2\cos\sqrt{1-\cos^2\theta}\Big\}$
$=\cos^{-1}\big\{2\cos\theta\sin\theta\big\}$
$\text{y}=\cos^{-1}\big\{\sin2\theta\big\}$
$\big[\text{Since}, \sin2\theta=2\sin\theta\cos\theta,\sin^2\theta+\cos^2\theta=1\big]$
$\text{y}=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)\Big]\ .....(\text{i})$
Now,
$\frac{1}{\sqrt{2}}<\text{x}<1$
$\Rightarrow\frac{1}{\sqrt{2}}<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0>-2\theta>-\frac{\pi}{2}$
$\Rightarrow\frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>0$
Hence, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\big[\text{Sicne}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=0-2\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2155 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}$
AnswerWe have, $\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{e}^{3\text{x}}+\log\sin4\text{x}+\log2^\text{x}$
$\Rightarrow\log\text{y}=3\text{x}\log\text{e}+\log\sin4\text{x}+\text{x}\log2$
$\Rightarrow\log\text{y}=3\text{x}+\log\sin4\text{x}+\text{x}\log2$
Differentiating with resepect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log2)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\log2(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\cot4\text{x}(4)+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+4\cot4\text{x}+\log2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[3+4\cot4\text{x}+\log2\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{3\text{x}}\sin4\text{x}2^\text{x}\big[3+4\cot4\text{x}+\log2\big]$
[Using equation (i)]
View full question & answer→Question 2165 Marks
Differentiate the following functions with respect to x:
$\sin(2\sin^{-1}\text{x})$
AnswerLet, $\text{y}=\sin(2\sin^{-1}\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sin(2\sin^{-1}\text{x})\Big)$
$=\cos\big(2\sin^{-1}\text{x}\big)\frac{\text{d}}{\text{dx}}\big(2\sin^{-1}\text{x}\big)$
[Using chain rule]
$=\cos\big(2\sin^{-1}\text{x}\big)\times2\frac{1}{\sqrt{1-\text{x}^2}}$
$=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\big(2\sin^{-1}\text{x}\big)\Big)=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2175 Marks
If $\text{x}=\text{a}\Big(\frac{1+\text{t}^2}{1-\text{t}^2}\Big)\text{ and y}=\frac{2\text{t}}{1-\text{t}^2},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\text{a}\Big(\frac{1+\text{t}^{2}}{1-\text{t}^{2}}\Big)$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})-(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})}\bigg]$[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})(2\text{t)}-(1+\text{t}^{2})(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{2\text{t}-2\text{t}^{3}+2\text{t}+2\text{t}^{3}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{4\text{a}\text{t}}{(1-\text{t}^{2})^{2}}\ .....(\text{i})$ and, $\text{y}=\frac{2\text{t}}{1-\text{t}^{2}}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(\text{t})-\text{t}\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})(1)-\text{t}(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})+2\text{t}^{2}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\ .....(\text{ii})$ Dividing equation (ii) by (i), $\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2(1+\text{t}^{2})}{(1-\text{t}^{2})^{2}}\times\frac{(1-\text{t}^{2})^{2}}{4\text{a}\text{t}}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{t}^{2})}{2\text{a}\text{t}}$
View full question & answer→Question 2185 Marks
If $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2},$ prove that $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
AnswerWe have, $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]$
$=\frac{\text{d}}{\text{dx}}\big(\text{x}\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^2}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$=\Big[\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big]-\frac{2\text{x}}{2\sqrt{1-\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}$
$=\sin^{-1}\text{x}$
View full question & answer→Question 2195 Marks
Differentiate the following functions with respect to x:
$\text{e}^{3\text{x}}\cos2\text{x}$
AnswerConsider $\text{y}=\text{e}^{3\text{x}}\cos2\text{x}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}\times\frac{\text{d}}{\text{dx}}(\cos2\text{x})+\cos2\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})$
[Using chain rule]
$=\text{e}^{3\text{x}}\times(-\sin2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x})+\cos2\text{xe}^{3\text{x}}\frac{\text{d}}{\text{dx}}(3\text{x})$
[Using chain rule]
$=-2\text{e}^{3\text{x}}\sin2\text{x}+3\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}}\cos2\text{x})=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
View full question & answer→Question 2205 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
AnswerLet, $\text{y}=\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\big)$
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\times\frac{1}{1+(\sqrt{\text{x}})^2}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{1+\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\Big)=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 2215 Marks
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
Taking log on both sides we get
$\text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}-\text{y}\tan\text{x}=\log\cos\text{y}-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}+\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\cos\text{x}+\text{x}\tan\text{y})=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}+\text{y}\tan\text{x}}{\log\cos\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 2225 Marks
If $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}},$ then find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}}$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\big]$
$=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\Big]$
$=\frac{1}{\sqrt{1-\big(6\text{x}\sqrt{1-9\text{x}^2}\big)^2}}\times\frac{\text{d}}{\text{dx}}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)$
$=\frac{1}{\sqrt{1-[36\text{x}^2(1-9\text{x}^2)]}}\times\Big(6\text{x}\frac{\text{d}}{\text{dx}}\sqrt{1-9\text{x}^2}+\sqrt{1-9\text{x}^2}\frac{\text{d}}{\text{dx}}(6\text{x})\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-9\text{x}^2)+\sqrt{1-9\text{x}^2}(6)\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}(-18\text{x}^2)+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2}{\sqrt{1-9\text{x}^2}}+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2+6\sqrt{1-9\text{x}^2}}{\sqrt{1-9\text{x}^2}}\Big)$
$=\frac{-54\text{x}^2+6-54\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
$=\frac{6-108\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
View full question & answer→Question 2235 Marks
If $\text{y}=\sqrt{\text{a}^2-\text{x}^2},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
AnswerDIfferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{a}^2-\text{x}^2}\big)$
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$
$=\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}$
$\big[\text{Since},\sqrt{\text{a}^2-\text{x}^2}=\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}$
Hence, the solution is, $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
View full question & answer→Question 2245 Marks
If $\text{x}-\text{e}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
$\text{y}=\text{e}^{\tan\text{x}\log\text{x}}+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\text{x})+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2}\log\Big(\frac{\text{x}^2+1}{2}\Big)\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{1}{2}\times\frac{2}{\text{x}^2+1}\times(\text{x})\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
View full question & answer→Question 2255 Marks
If $\text{xy}=1,$ prove that $\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
AnswerHere, xy = 1 .....(i)
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$ \Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)=0$
$\Big[\text{Put x}=\frac{1}{\text{y}}\text{ from equation (i)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\frac{1}{\text{y}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
View full question & answer→Question 2265 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
Put $\text{x}=\text{a}\tan\theta$
$\Rightarrow\text{y}=\sin^{-1}\Big\{\frac{\text{a}\tan\theta}{\sqrt{\text{a}^2\tan^2\theta+\text{a}^2}}\Big\}$
$\Rightarrow\text{y}=\sin^{-1}\bigg\{\frac{\text{a}\tan\theta}{\sqrt{\text{a}^2(\tan^2\theta+1)}}\bigg\}$
$\Rightarrow\text{y}=\sin^{-1}\Big(\frac{\text{a}\tan\theta}{\text{a}\sec\theta}\Big)$
$\Rightarrow\text{y}=\sin^{-1}(\sin\theta)$
$\Rightarrow\text{y}=\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since, x} = \text{a}\tan\theta\big]$
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 2275 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\cos^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\cos^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\cos^{-1}\text{x}}$
$\Rightarrow\log\text{y}=\cos^{-1}\text{x}\log\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos^{-1}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\cos^{-1}\text{x}}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 2285 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\}$
Put $\text{x}=\sin\theta$
$\text{y}=\tan^{-1}\Big\{\frac{\sin\theta}{1+\sqrt{1-\sin^2\theta}}\Big\}$
$\text{y}=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$\text{y}=\tan^{-1}\bigg\{\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\bigg\}$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\sin^{-1}\text{x}\big[\text{Since, x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2295 Marks
Differentiate $\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$ with respect to $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ if $-1<\text{x}<1,\text{x}\neq0.$
AnswerLet $\text{u}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$
Put $\text{x}=\tan\theta, \text{So}$
$\text{u}=\tan^{-1}\Big(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\sin\theta}{2}}{\frac{\cos\theta}{2}}\bigg)$
$\text{u}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)\ .....(\text{i})$
And,
Let, $\text{v}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-1<\text{x}<1$
$\Rightarrow -1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}\ .....(\text{A})$
So, from equation (i),
$\text{u}=\frac{\theta}{2}\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\frac{1}{2}\tan^{-1}\text{x}\big[\text{Since},\text{x}=\tan\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{2}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{du}}{\text{dx}}=\frac{1}{2}\frac{1}{(1+\text{x}^2)}\ .....(\text{iii})$
Now, from equation (ii) and (A)
$\text{v}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{v}=2\tan^{-1}\text{x}\big[\text{Since},\text{x}=\tan\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=2\Big(\frac{1}{1+\text{x}^2}\Big)\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{2(1+\text{x}^2)}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{4}$
View full question & answer→Question 2305 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta),\text{y}=\text{ae}^\theta(\sin\theta+\cos\theta)$
AnswerWe have, $\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta)$ and $\text{y}=\text{ae}^{\theta}(\sin\theta+\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta-\cos\theta)+(\sin\theta-\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta+\cos\theta)+(\sin\theta+\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta+\sin\theta)+(\sin\theta-\cos\theta)(\text{e}^\theta)\Big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta-\sin\theta)+(\sin\theta+\cos\theta)(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\sin\theta)\big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\cos\theta)\big]$
$\therefore\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}(2\text{e}^\theta\cos\theta)}{\text{a}(2\text{e}^\theta\sin\theta)}=\cot\theta$
View full question & answer→Question 2315 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{\text{e}^\text{t}+\text{e}^{-\text{t}}}{2}\text{ and y}=\frac{\text{e}^\text{t}-\text{e}^\text{-t}}{2}$
AnswerWe have, $ \text{x}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$ and $ \text{y}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})+\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})-\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}+\text{e}^{\text{-t} \frac{\text{d}}{\text{dt}}}(\text{-t})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}-\text{e}^{\text{-t}}\frac{\text{d}}{\text{dt}}({\text{e}^{\text{-t}}})\bigg]$
$\Rightarrow\frac{1}{2}(\text{e}^{\text{t}}-\text{e}^\text{-t})=\text{y}$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}(\text{e}^\text{t}+\text{e}^{\text{-t}})=\text{x}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{x}}{\text{y}}$
View full question & answer→Question 2325 Marks
If $\text{xy}=4,$ prove that $\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
AnswerWe have, $\text{xy}=4$
$\Rightarrow\text{y}=\frac{4}{\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\frac{4}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4(-1\times\text{x}^{-1-1})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4}{\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4}{\big(\frac{4}{\text{y}}\big)^2}\ \Big[\because\text{x}=\frac{4}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4\text{y}^2}{16}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2}{4}$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}+4\text{y}^2=3\text{y}^2$
$\Rightarrow4\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}^2$
Dividing both side by x,
$\Rightarrow\frac{4}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
View full question & answer→Question 2335 Marks
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
AnswerWe have, $\text{y}=\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$
View full question & answer→Question 2345 Marks
If $x^m + y^n = 1$, Prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
AnswerWe have, $x^m + y^n = 1$
Taking log on both side,
$\log(\text{x}^\text{m}\text{y}^{\text{n}})=\log(1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}(\text{m}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{n}\log\text{y})=\frac{\text{d}}{\text{dx}}\big\{\log(1)\big\}$
$\Rightarrow\frac{\text{m}}{\text{n}}+\frac{\text{n}}{\text{y}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{m}}{\text{x}}\times\frac{\text{y}}{\text{n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
View full question & answer→Question 2355 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\sin\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\sin\text{x}}(\cos\text{x})-\sin\text{x}\log\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
View full question & answer→Question 2365 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\tan^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\tan^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\tan^{-1}\text{x}}$
$\log\text{y}=\tan^{-1}\text{x}\log\text{x}\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\tan^{-1}\text{x}}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 2375 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:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\cos\theta\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dv}}{\text{dx}}=2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{-2}{\sqrt{1-4\text{x}^2}}(2)\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (v) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{x}}$
View full question & answer→Question 2385 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\}$ Put $\text{x}=\text{a}\sin\theta$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2-\text{a}^2\sin^2\theta}}\Big\}$ $\text{y}=\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}\cos\theta}\Big\}$ $\text{y}=\tan^{-1}(\tan\theta)\ .....(\text{i})$ Here, $-\text{a}<\text{x}<\text{a}$ $\Rightarrow-1<\frac{\text{x}}{\text{a}}<1$ $\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$ From equation (i), $\text{y}=\theta$ $\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$ $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since x}=\text{a }\sin \theta\big]$ Differentiating it with respect to x, $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$ $=\frac{\text{a}}{\sqrt{\text{a}^2-\text{x}^2}}\times\big(\frac{1}{\text{a}}\big)$$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}$
View full question & answer→Question 2395 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big), -\frac{1}{\sqrt{6}}<\text{x}<\frac{1}{\sqrt{6}}$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big)$
$=\tan^{-1}\Big(\frac{3\text{x}+2\text{x}}{1-(3\text{x})(2\text{x})}\Big)$
$\text{y}=\tan^{-1}(3\text{x})+\tan^{-1}(2\text{x})$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{1+(2\text{x})^2}\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{1}{1+9\text{x}^2}(3)+\frac{1}{1+4\text{x}^2}(2)$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}+\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 2405 Marks
If $\text{x}=\text{a}(\theta+\sin\theta),\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{x}=\text{a}(\theta+\sin\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big(\frac{\text{d}}{\text{d}\theta}(\theta)+\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta)\ .....(\text{i})$
And, $\text{y}=\text{a}(1+\cos\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(0-\sin\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sin\theta\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1+\cos\theta)}$
$=\frac{-\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\tan\theta}{2}$
View full question & answer→Question 2415 Marks
If $\text{x}=\cos\text{t}(3-2\cos^2\text{t}),\text{y}\sin\text{t}(3-2\sin^2\text{t})$ find the value of $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{4}$
Answer$\text{x}=\cos\text{t}(3-2\cos^2\text{t})\text{ and }\text{y}\sin\text{t}(3-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}(3-2\cos^2\text{t})+\cos\text{t}(4\cos\text{t}\sin\text{t})$ and $\frac{\text{dy}}{\text{dx}}=\cos\text{t}(3-2\sin^2\text{t})+\sin\text{t}(-4\sin\text{t}\cos\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+6\sin\text{t}\cos^2\text{t}$ and $ \frac{\text{dy}}{\text{dx}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+(1-2\cos^2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\sin\text{t}\cos2\text{t}\cos2\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\cos\text{t}\cos2\text{t}}{3\sin\text{t}\cos2\text{t}}=\cot\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1$
View full question & answer→Question 2425 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$x^5 + y^5 = 5xy$
AnswerWe Heve, $x^5 + y^5 = 5xy$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^5\big)+\frac{\text{d}}{\text{dx}}\big(\text{y}^5\big)=\frac{\text{d}}{\text{dx}}\big(5\text{xy}\big)$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{dy}}{\text{dx}}\big(\text{x}\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\big(1\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\text{x}\frac{\text{dy}}{\text{dx}}+5\text{y}$
$\Rightarrow5\text{y}^4\frac{\text{dy}}{\text{dx}}-5\text{x}\frac{\text{dy}}{\text{dx}}=5\text{y}-5\text{x}^4$
$\Rightarrow5\frac{\text{dy}}{\text{dx}}\big(\text{y}^4-\text{x}\big)=5\big(\text{y}-\text{x}^4\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{5(\text{y}-\text{x}^4)}{5(\text{y}^4-\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}^4}{\text{y}^4-\text{x}}$
View full question & answer→Question 2435 Marks
Differentiate $\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$ with respect to x:
AnswerWe have $\text{y}=\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\Big\{\frac{2\times6^\text{x}}{1+6^{2\text{x}}}\Big\}$
Put $6^\text{x}=\tan\theta$
$\Rightarrow \theta=\tan^{-1}(6^\text{x})$
Now, $\text{y}=\sin^{-1}\Big\{\frac{2\tan\theta}{1+\tan^2\theta}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\big\{\sin2\theta\big\}$
$\Rightarrow \text{y}=2\theta$
$\Rightarrow \text{y}=2\tan^{-1}(6^\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\times\frac{1}{(6^\text{x})^2}\times6^\text{x}\log6$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2(\log6)6^\text{x}}{36^\text{x}}$
View full question & answer→Question 2445 Marks
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
AnswerConsider $\text{y}=\cos(\log\text{ x})^2$
Differentiating it with respect to x and applying the chain and the product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}+\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}$
$=\frac{1}{2}(\text{x}+1)^\frac{-1}{2}+\frac{1}{2}(\text{x}-1)^\frac{-1}{2}$
$=\frac{1}{2}\Big(\frac{1}{\sqrt{\text{x}+1}}\frac{1}{\sqrt{\text{x}-1}}\Big)$
$=\frac{1}{2}\bigg(\frac{\sqrt{\text{x}-1}+\sqrt{\text{x}+1}}{\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{y}}{\sqrt{\text{x}^2-1}}\Big)$
So,
$\sqrt{\text{x}^2}-\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}$
View full question & answer→Question 2455 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\cos^{-1}\Big\{\sqrt{\frac{2\cos^2\theta}{2}}\Big\}$
$\text{y}=\cos^{-1}\{\cos\theta\}$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\pi$
$\Rightarrow\ 0<\theta<\frac{\pi}{2}$
So, equation (i),
$\text{y}=\theta$
$\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→