Question 1015 Marks
If $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^\text{a},\text{y}=\text{a}^{\text{t}+\frac{1}{\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere, $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\left[\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}\right]$
$=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\Big(\text{1}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}^{(\text{t}+\frac{1}{\text{t}})}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\bigg[\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\bigg]$
$=\frac{\text{d}}{\text{dt}}\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\frac{\text{d}}{\text{dt}}\left(\text{t}+\frac{1}{\text{t}}\right)$
$\frac{\text{dy}}{\text{dt}}=\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\left(1-\frac{1}{\text{t}^{2}}\right)}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}\big(1-\frac{1}{\text{t}}^{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}}$
View full question & answer→Question 1025 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}$
AnswerHere,
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}\ .....(\text{i})$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{ax}}\times\sec^\text{x}\times\log\text{x}}{(1-2\text{x})^\frac{1}{2}}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{\text{ax}}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \begin{bmatrix} \text{Since}, \log\Big(\frac{\text{A}}{\text{B}}\Big)=\log\text{A}-\log\text{B},\\ \log(\text{AB})=\log\text{A}+\log\text{B} \end{bmatrix}$
$\log\text{y}=\text{ax}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a and }\log_\text{e}\text{e}=1\big]$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{ax})+\frac{\text{d}}{\text{dx}}(\log\sec\text{x})+\frac{\text{d}}{\text{dx}}(\log\log\text{x})-\frac{1}{2}\log(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{1}{\sec\text{x}}\frac{\text{d}}{\text{dx}}(\sec\text{x})+\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)\frac{\text{d}}{\text{dx}}(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{\sec\text{x}\tan\text{x}}{\sec\text{x}}+\frac{1}{(\log\text{x})}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)(-2)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{ax}}\sec\text{x}\log\text{x}}{\sqrt{1-2\text{x}}}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 1035 Marks
If $x^{13} y^7 = (x + y)^{20}$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerHere,
$x^{13}y^7 = (x + y)^{20}$
Taking log on both the sides,
$\log(\text{x}^{13}\text{y}^7)=\log(\text{x}+\text{y})^{20}$
$13\log\text{x}+7\log\text{y}=20\log(\text{x}+\text{y})$
$\big[\text{Since},\log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule,
$13\frac{\text{d}}{\text{dx}}(\log\text{x})+7\frac{\text{d}}{\text{dx}}(\log\text{y})=20\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{20}{(\text{x}+\text{y})}=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{7}}{\text{y}}-\frac{20}{(\text{x}+\text{y})}\Big]=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{2(\text{x}+\text{y})-20\text{y}}{\text{y}(\text{x}+\text{y})}\Big]=\Big[\frac{20\text{x}-13(\text{x}+\text{y})}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{20\text{x}-13\text{x}-13\text{y}}{\text{x}(\text{x}+\text{y})}\Big]\Big(\frac{\text{y}(\text{x}+\text{y})}{7\text{x}+7\text{y}-20\text{y}}\Big)$
$=\frac{\text{y}}{\text{x}}\Big(\frac{7\text{x}-13\text{y}}{7\text{x}-13\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 1045 Marks
If $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}\frac{(1-\tan\text{a})}{(1+\tan\text{a})}$
AnswerWe have, $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\tan\text{a}$
$\Rightarrow\text{x}^2-\text{y}^2=\tan\text{a}(\text{x}^2+\text{y}^2)$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{y}^2)=\tan\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)$
$\Rightarrow\Big(2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\tan\text{a}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}\tan\text{a}+2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}-2\text{x}\tan\text{a}$
$\Rightarrow2\text{y}(1+\tan\text{a})\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\tan\text{a})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\Big(\frac{1-\tan\text{a}}{1+\tan\text{a}}\Big)$
View full question & answer→Question 1055 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
AnswerLet $\text{y}=\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\log(\tan\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\text{x}\log(\tan\text{x})}$
Differentiating with resepect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{x}\log(\tan\text{x})}\big\}$
$=\text{e}^{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\text{e}^{\text{x}\log(\tan\text{x})}\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{x})$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+\text{e}^{\log(\tan\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\Big[\frac{\text{x}}{\tan\text{x}}(\sec^2\text{x})+\log\tan\text{x}\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\big[\text{x}\sec\text{x cosec x}+\log\tan\text{x}\big]$
View full question & answer→Question 1065 Marks
If $\text{y}=\cot^{-1}\Big\{\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big\},$ show that $\frac{\text{dy}}{\text{dx}}$ is independent of x.
AnswerLet $\text{y}=\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big] \ .....(\text{i})$
Then, $\frac{\sqrt{1+\sin\theta}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}$
$=\frac{\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)^2}{\big(\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}\big)\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)} $
$=\frac{(1+\sin\text{x})+(1-\sin\text{x})+2\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{(1+\sin\text{x})-(1-\sin\text{x})}$
$=\frac{2+2\sqrt{1-\sin^2\text{c}}}{2\sin\text{x}}$
$=\frac{1+\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\cot\frac{\text{x}}{2}$
Therefore, equation (i) becomes
$\text{y}=\cot^{-1}\Big(\cot\frac{\text{x}}{2}\Big)$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 1075 Marks
If $(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\tan\text{y}-\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}$
AnswerHere,
$(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\cos\text{x})^{\text{y}}=\log(\tan\text{y})^{\text{x}}$
$\text{y}\log(\cos\text{x})=\text{x}\log(\tan\text{y})$
$\big[\text{Since}, \log\text{e}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{y})$
$\Big(\text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{y}+\log\tan\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big)$
$\Big(\text{y}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{1}{\tan\text{y}}\frac{\text{d}}{\text{dx}}(\tan\text{y})+\log\tan\text{y}(1)\Big)$
$\Big(\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big)\\ =\Big(\frac{\text{x}}{\tan\text{y}}(\sec^2\text{y})\Big)\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}-\text{y}\tan\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}} \\ =\Big(\sec\text{y cosec y}\times\text{y}\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}\Big)$
$\frac{\text{dy}}{\text{dx}}\big[\log\cos\text{x}-\text{x}\sec\text{y cosec y}\big] \\ =\log\tan\text{y}+\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{\log\tan\text{x}+\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}\Big]$
View full question & answer→Question 1085 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
AnswerHere,
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}+\cos^2\big(\frac{\pi}{2}\big)$
$\text{y}=\text{e}^{\log\text{x}\log\tan\text{x}}+\cos\text{x}^2\big(\frac{\pi}{4}\big)$
$\big[\text{Since, e}^{\log\text{a}}=\text{a and}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating ti using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\log\text{x}\log\tan\text{x}})+\frac{\text{d}}{\text{dx}}\cos^2\big(\frac{\pi}{4}\big)$
$=\text{e}^{\log\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x}\log\tan\text{x})+0$
$=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}\Big[\log\times\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}\Big(\frac{1}{\text{x}}\Big)\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\big(\sec^2\text{x}\big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{\sec^2\text{x}}{\tan\text{x}}\Big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
View full question & answer→Question 1095 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big)$
AnswerLet, $\text{y}=\tan^{-1}\Big[\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}}}{\frac{\text{b}-\text{a}\tan\text{x}}{\text{b}}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}}{\text{b}}+\tan\text{x}}{1-\frac{\text{a}}{\text{b}}\tan\text{x}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Bigg[\frac{\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)+\tan\text{x}}{1-\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)\times\tan\text{x}}\Bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Big[\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}+\text{x}\Big)\Big]$
$\Rightarrow\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\therefore\ \frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 1105 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\log\text{x}}$
$\Rightarrow\log\text{y}=\log\text{x}\log\sin\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{\sin\text{x}}(\cos\text{x})+\frac{\log\sin\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\log\text{x}}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 1115 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy}\log(\text{x}+\text{y})=1\ .....(\text{i})$
Differentiaitng with respect to x using chain rula, product rule,
$\frac{\text{dy}}{\text{dx}}(\text{xy}\log(\text{x}+\text{y}))=\frac{\text{d}}{\text{dx}}(1)$
$\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\frac{\text{xy}}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{x}\log(\text{x}+\text{y})=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\frac{\text{xy}}{\text{x}+\text{y}}+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
View full question & answer→Question 1125 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^\text{x}$
AnswerLet $\text{y}=(\log\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log(\log\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x, using product rule, chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$=\frac{\text{x}}{\log\text{x}}\Big(\frac{1}{\text{x}}\Big)+\log\log\text{x}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}+\log\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
[Using equation (i)]
View full question & answer→Question 1135 Marks
If $(\sin\text{x})^{\text{y}}=\text{x}+\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-(\text{x}+\text{y})\text{y}\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}$
AnswerHere,
$(\sin\text{x})^{\text{y}}=\text{x}+\text{y}$
Taking log on both the sides,
$\log(\sin\text{x})^\text{y}=\log(\text{x}+\text{y})$
$\text{y}\log(\sin\text{x})=\log(\text{x}+\text{y})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log(\sin\text{x}))=\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\text{y}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{\text{y}}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{\text{y}(\cos\text{x})}{(\sin\text{x})}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\log\sin\text{x}-\frac{1}{\text{x}+\text{y}}\Big)=\frac{1}{(\text{x}+\text{y})}-\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{(\text{x}+\text{y})\log\sin\text{x}-1}{(\text{x}+\text{y})}\Big)=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{\text{x}+\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}\Big)$
View full question & answer→Question 1145 Marks
If $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}},\frac{\pi}{4}<\text{x}<\frac{3\pi}{4},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\log\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x}) \\ +(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})(\sin\text{x}-\cos\text{x}) \\ +\frac{(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})}\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})} \\ \big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
View full question & answer→Question 1155 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
AnswerLet, $\text{y}=\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big]$
$=\Bigg[\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{x}}-\text{e}^{4\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\frac{(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}})\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})-\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]}{\big(\text{e}^{2\text{x}}-\text{e}^{2\text{x}}\big)^2}$
$=\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{x}}-2\text{e}^{-2\text{e}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{e}}+2\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{e}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2-2\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)^2}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big[\text{e}^{4\text{x}}+\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}-\text{e}^{4\text{x}}-\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}\big]}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big)=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
View full question & answer→Question 1165 Marks
If $\text{x}=\text{e}^{\cos2\text{t}}$ and $\text{y}=\text{e}^{\sin2\text{t}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
AnswerWe have, $\text{x}=(\text{e}^{\cos2\text{t}})$ and $\text{y}=\text{e}^{\sin2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\cos2\text{t}}\big)$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\sin2\text{t}}\big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$ and $\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}(-\sin2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=\text{e}^{\sin2\text{t}}(\cos2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-2\sin 2\text{t}\text{e}^{\cos2\text{t}}$ and $\frac{\text{dy}}{\text{dt}}=2\cos 2\text{t}\text{e}^{\sin2\text{t}}$
$\because\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\cos2\text{te}^{\sin2\text{t}}}{-2\sin2\text{te}^{\cos2\text{t}}}$
$\begin{bmatrix} \because\text{x}=\text{e}^{\cos2\text{t}}\Rightarrow\log\text{x}=\cos2\text{t} \\ \text{y}=\text{e}^{\sin2\text{t}}\Rightarrow\log\text{y}=\sin2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
View full question & answer→Question 1175 Marks
If $e^y= y^x$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{\log\text{y}-1}$
AnswerWe have,$ e^y = y^x$
Taking log on both sides,
$\log\text{e}^{\text{y}}=\log\text{y}^\text{x}$
$\Rightarrow\text{y}\log\text{e}=\text{x}\log\text{y}$
$\Rightarrow\text{y}=\text{x}\log\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(1-\frac{\text{x}}{\text{y}}\Big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\frac{\text{y}-\text{x}}{\text{y}}\big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{y}\log\text{y}}{\Big(\text{y}-\frac{\text{y}}{\log\text{y}}\Big)}$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}(\log\text{y})}{\text{y}\log\text{y}-\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\log\text{y})^2}{\text{y}(\log\text{y}-1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{(\log\text{y}-1)}$
View full question & answer→Question 1185 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = x^n + n^x + x^x + n^n$
AnswerWe have, $y = x^n + n^x + x^x + n^x$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\log\text{x}^\text{x}}+\text{n}^\text{n}$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\text{x}\log\text{x}}+\text{n}^\text{n}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{n})$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{e}^{\log\text{x}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\Big[\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[1+\log\text{x}\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[\log\text{e}+\log\text{x}\big] \\ \big[\because\log_\text{e}\text{e}=1\text{ and }\log\text{A}+\log\text{B}=\log(\text{AB})\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\log\big(\text{ex}\big)$
View full question & answer→Question 1195 Marks
If $\text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)\text{ and y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
AnswerHere, $ \text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
Differentiating it with respect to t,
$\frac{\text{dx}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}-1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
Differetiating it with respect to t,
$\frac{\text{dy}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(1+\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dy}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\text{a}\frac{(\text{t}^{2}+1)}{\text{t}^{2}}\times\frac{\text{t}^{2}}{\text{a}(\text{t}^{2}-1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{t}^{2}+1}{\text{t}^{2}-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
$\Big[\text{Since},\frac{\text{x}}{\text{y}}=\frac{\text{a}(\text{t}^{2}+1)}{\text{t}}\times\frac{\text{t}}{\text{a}(\text{t}^{2}-1)}=\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}-1}\Big)\Big]$
View full question & answer→Question 1205 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\theta=\sin^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}(\tan\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
$\text{v}=\sin^{-1}(2\sin\theta\sqrt{1-\sin^{2}\theta})$
$\text{v}-\sin^{-1}(2\sin\theta\cos\theta)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i)
$\text{u}=\theta\Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\sin^{-1}\text{x}$
Differentiatiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}}{2}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{1}{2}$
View full question & answer→Question 1215 Marks
Differentiate $\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$ with respect to $\sqrt{1-\text{x}^2},$ if -1 < x < 1.
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta$
$\Rightarrow\theta=\tan^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]\ .....(\text{i})$
Here,
$-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow\frac{\pi}{4}>-\theta>\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{4}<-\theta<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}-\theta<\frac{\pi}{2}$
So, from equation (i)
$\text{u}=\frac{\pi}{4}-\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\tan^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{1}{1+\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=-\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
And
Let, $\text{v}=\sqrt{1-\text{x}^2}$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\times\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
Dividing equation (ii) by (iii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x} ^2}}{-\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{\text{x}(1+\text{x}^2)}$
View full question & answer→Question 1225 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet, $\text{y}=\text{e}^\sqrt{{\cot\text{x}}}$
$\Rightarrow\ \text{y}=\text{e}^{(\cot\text{x})^\frac{1}{2}}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{(\cot\text{x})^\frac{1}{2}}\Big)$
$=\text{e}^{(\cot\text{x})^\frac{1}{2}}\frac{\text{d}}{\text{dx}}(\cot\text{x})^\frac{1}{2}$
[Using chain rule]
$=\text{e}^\sqrt{\cot\text{x}}\times\frac{1}{2}(\cot\text{x})^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}(\cot\text{x})$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
View full question & answer→Question 1235 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:
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 1245 Marks
If $\text{x}=\cot\text{t and y}=\sin\text{t},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{3}}\text{ at t}=\frac{2\pi}{3}$
AnswerWe have, $\text{x}=\cos\text{t}$ and $\text{y}=\sin\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos\text{t}) $ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t}) $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\cos\text{t}$
$\therefore\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{-\sin\text{t}}=-\cot{\text{t}}$
Now, $\big(\frac{\text{dy}}{\text{dx}}\big)_{\text{t}=\frac{2\pi}{3}}=-\cot\big(\frac{2\pi}{3}\big)=\frac{1}{\sqrt{3}}$
View full question & answer→Question 1255 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
$\Rightarrow\ \text{y}=\tan^{-1}\Big(\frac{3\text{x}-2\text{x}}{1+(3\text{x})(2\text{x})}\Big)$
$\Rightarrow\text{y}=\tan^{-1}3\text{x}-\tan^{-1}2\text{x}$
$\Big[\text{Since}, \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})-\frac{1}{1+(2\text{x})^3}\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+9\text{x}^2}(3)-\frac{1}{1+4\text{x}^2}(2)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}-\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 1265 Marks
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerWe have, $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}+\text{y}}{\text{x}-\text{y}}=\sec^{-1}({\text{a}})$
Differentiate with respect to x, we get,
$\Rightarrow\bigg[\frac{(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})-(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})}{(\text{x}-\text{y}}\bigg]=0$
$\Rightarrow(\text{x}-\text{y})\Big(1+\frac{\text{d}}{\text{dx}}\Big)-(\text{x}+\text{y})\Big(1-\frac{\text{d}}{\text{dx}}\Big)=0$
$\Rightarrow(\text{x}-\text{y})+(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}-(\text{x}+\text{y})+(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[\text{x}-\text{y}+\text{x}+\text{y}]=\text{x}+\text{y}-\text{x}+\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{x})=2\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 1275 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(-1,0)$
AnswerLet $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta, \text{So},$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Here,
$\text{x}\in(-1,0)$
$\Rightarrow\cos\theta\in(-1, 0)$
$\Rightarrow\theta\in\Big(\frac{\pi}{2},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\theta\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)\Big]$
$\text{u}=\pi-\cos^{-1}\text{x }\big[\text{Since}\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, 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)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=-1$
View full question & answer→Question 1285 Marks
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
AnswerHere, $\text{y}=\sqrt{\text{x}^2+\text{a}^2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+\text{a}^2}\big)$
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{a}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\times(2\text{x})$
$=\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\ \Big[\text{Since},\sqrt{\text{x}^2+\text{a}^2}=\text{y}\Big]$
$\Rightarrow \text{y}\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
View full question & answer→Question 1295 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
View full question & answer→Question 1305 Marks
If $\text{y}=\text{e}^{\text{x}}+\text{e}^{-\text{x}},$ prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
AnswerDifferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)$
$=\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}+\frac{\text{d}}{\text{dx}}{\text{e}}^{-\text{x}}$
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\big(-\text{x}\big)$
[Using chain rule]
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}(-1)$
$=\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)$
$=\sqrt{\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)^2-4\text{e}^{\text{x}}\times\text{e}^{-\text{x}}}$
$\Big[\text{Since},(\text{a}-\text{b}=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}\Big]$
$=\sqrt{\text{y}^2-4}$
$\big[\text{Since e}^\text{x}+\text{e}^{-\text{x}}=\text{y}\big]$
Hence, the solution is, $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
View full question & answer→Question 1315 Marks
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
AnswerHere,
$\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\sin\text{x})^\text{y}$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\text{y}}$
$\log\text{y}=\text{y}(\log\sin\text{x})$
Differentiating it with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\sin\text{x}\Big)=\frac{\text{y}}{\sin\text{x}}(\cot\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\text{y}\log\sin\text{x}}{\text{y}}\Big)=\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}^2\cot\text{x}}{1-\text{y}\log\sin\text{x}}\Big)$
View full question & answer→Question 1325 Marks
Differentiate the following functions with respect to x:
$(\tan\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\tan\text{x})^\frac{1}{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\tan\text{x})^\frac{1}{\text{x}}$
$\log\text{y}=\frac{1}{\text{x}}\log(\tan\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\log(\tan\text{x})+\log(\tan\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{\text{x}}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log(\tan\text{x})\Big(-\frac{1}{\text{x}^2}\Big)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\tan\text{x}}(\sec^2\text{x})-\frac{\log(\tan\text{x})}{\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^\frac{1}{\text{x}}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 1335 Marks
Differentiate the following functions with respect to x:
$\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
AnswerLet $\text{y}=\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dy}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Bigg[\frac{(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)-(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)}{(1-\text{x})^2}\Bigg]$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\bigg[\frac{(1-\text{x}^2)(2\text{x})-(1+\text{x}^2)(-2\text{x})}{(1-\text{x}^2)^2}\bigg]$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big[\frac{2\text{x}-2\text{x}^3+2\text{x}+2\text{x}^3}{(1-\text{x}^2)^2}\Big]$
$=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
View full question & answer→Question 1345 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)\Big\}$
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)$
[Using chain rule and quotient rule]
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)^2}}\times\Bigg[\frac{(\text{x}^2+\text{a}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)^\frac{1}{2}}{\Big[(\text{x}^2+\text{a}^2)^\frac{1}{2}\Big]^2}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\sqrt{\text{x}^2+\text{a}^2-\text{x}^2}}\times\Bigg[\frac{\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)}{(\text{x}^2+\text{a}^2)}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\times2\text{x}\Big]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\frac{\text{x}^2+\text{a}^2-\text{x}^2}{\sqrt{\text{x}^2+\text{a}^2}}\Big]$
$=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
$=\frac{\text{a}}{(\text{x}^2+\text{a}^2)}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)\Big\}=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
View full question & answer→Question 1355 Marks
If $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
AnswerConsider the given function, $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a}.$
We need to prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}.$
Differentiating the given equation w.r.t 'x' we get
$(\text{x}-\text{y})\Bigg[\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}\Bigg(\frac{(\text{x}-\text{y})-\text{x}\big(1-\frac{\text{dy}}{\text{dx}}\big)}{(\text{x}-\text{y})^2}\Bigg)\Bigg]+\text{e}^\frac{\text{x}}{\text{x}-\text{y}}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\frac{(\text{x}-\text{y})-\text{x}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)}{(\text{x}-\text{y})}+\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(1-\frac{\text{x}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(\frac{-\text{y}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow-\text{y}+\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}-\text{y}=0$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
View full question & answer→Question 1365 Marks
If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos\text{t}-\cos3\text{t}$
$\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}$
$\frac{\text{dy}}{\text{dt}}=-3\sin\text{t}+3\sin3\text{t}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
When $\text{t}=\frac{\pi}{3}$
$\frac{\text{dy}}{\text{dx}}=\frac{-3\sin\big(\frac{\pi}{3}\big)+3\sin(\pi)}{3\cos\big(\frac{\pi}{3}\big)-3\cos(\pi)}=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt{3}}$
View full question & answer→Question 1375 Marks
Differentiate the following functions from first principles:$e^{ax+b}.$
AnswerLet $f(x) = e^{ax+b}$
$\Rightarrow f(x + h) = e^{a(x+h)+b}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}(\text{x}+\text{h})+\text{b}}-\text{e}^{(\text{ax}+\text{b})}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}+\text{b}}\text{e}^{\text{ah}}-\text{e}^{\text{ax}+\text{b}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\text{ax}+\text{b}}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}\times\text{a}$
$=\text{ae}^{\text{ax}+\text{b}} \lim\limits_{\text{h}\rightarrow0}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}$
$=\text{ae}^{\text{ax}+\text{b}}$ So, $\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}+\text{b})$
$=\text{ae}^{\text{ax}+\text{b}}$
View full question & answer→Question 1385 Marks
If $\tan(\text{x}+\text{y})+\tan(\text{x}+\text{y})=1,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\tan(\text{x}+\text{y})+\tan(\text{x}-\text{y})=1$
Differentiating with respect to x, we get,
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})+\frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]+\sec^2(\text{x}-\text{y})\Big[1-\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\sec^2(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}} \\ =-\big[\sec^2(\text{x}-\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\sec^2(\text{x}+\text{y})-\sec^2(\text{x}-\text{y})\big] \\ =-\big[\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})}{\sec^2(\text{x}-\text{y})-\sec^2(\text{x}+\text{y})}$
View full question & answer→Question 1395 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
AnswerWe have, $\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log\text{x}^{\cos\text{x}}}+\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}$
$\text{y}=\text{e}^{\cos\text{x}\log\text{x}}+\text{e}^{\tan\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x}\log\text{x}) \\ +\text{e}^{\tan\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\sin\text{x})$
$=\text{e}^{\log\text{x}^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}\Big[\tan\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\cos\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(-\sin\text{x})\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin)\text{x}+\log\sin\text{x}(\sec^2\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\sec^2\text{x}\log\sin\text{x}\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\big[1+\sec^2\text{x}\log\sin\text{x}\big]$
View full question & answer→Question 1405 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here, $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}= 2\theta\bigg[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]\bigg]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\bigg[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\bigg]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}[\text{since, x}=\sin\theta]$
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 1415 Marks
If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big)$
Put $\text{x}=\cos2\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2}(\cos\theta-\sin\theta)}{\sqrt{2}(\cos\theta+\sin\theta)}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta-\sin\theta}{\cos\theta}}{\frac{\cos\theta+\sin\theta}{\cos\theta}}\bigg)$
[Dividing numerator and denomainator by $\cos\theta$]
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}\bigg)$
$=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\times\tan\theta}\bigg)$
$=\tan^{-1}\Big[\tan\big(\frac{\pi}{4}-\theta\big)\Big]$
$=\frac{\pi}{4}-\theta$
$=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\ (\text{Using x}=\cos2\theta)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1425 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
AnswerLet $\text{y}=\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
$\Rightarrow\ \text{y}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{-1}{2}}\bigg[\frac{(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})-(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})}{(1-\text{x})^2}\bigg]$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}}\bigg[\frac{(1-\text{x})(1)-(1+\text{x})(-1)}{(1-\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^{\frac{1}{2}}\bigg[\frac{1-\text{x}+1+\text{x}}{(1-\text{x})^2}\bigg]$
$=\frac{1}{2}\frac{1+\text{x}^\frac{1}{2}}{1-\text{x}^\frac{1}{2}}^{\frac{1}{2}}\times\frac{1}{(1-\text{x})^2}$
$=\frac{1}{\sqrt{1+\text{x}}(1-\text{x})^\frac{3}{2}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big)=\frac{1}{\sqrt{1+\text{x}}(1-\text{x})^\frac{3}{2}}$
View full question & answer→Question 1435 Marks
Differentiate the following functions with respect to x:
$\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
AnswerLet, $\text{y}=\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
$\Rightarrow\ \text{y}=\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{-1}{2}}\times\frac{1}{1+\big(\frac{\text{x}}{2}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{2}\big)$
$=\frac{4}{4\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\big(4+\text{x}^2\big)}$
$=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\bigg)=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
View full question & answer→Question 1445 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{1-\sqrt{\text{xa}}}\Big)$$\text{y}=\tan^{-1}\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{a}}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{x}})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\sqrt{\text{a}})$
$=\frac{1}{1+\big(\sqrt{\text{x}}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+0$
$=\Big(\frac{1}{1+\text{x}}\Big)\Big(\frac{1}{2\sqrt{\text{x}}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 1455 Marks
If $\sin^2\text{y}+\cos\text{xy}=\text{k},$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1,\text{y}=\frac{\pi}{4}$
AnswerHere, $\text{e}^{\text{x}}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
Differentiating with respect to x using chain rule,
$\Rightarrow \frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\text{e}^{\text{y}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}+\text{y}}\big)$
$\Rightarrow \text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dy}}(\text{a}+\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}}\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{dy}}{\text{dx}}\big(\text{e}^{\text{y}}-\text{e}^{\text{x}+\text{y}}\big)=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{e}^\text{x}\times\text{e}^\text{y}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}\times\text{e}^\text{y}}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({1-\text{e}}^\text{x}\big)}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({\text{e}^\text{x}-1}\big)}$
View full question & answer→Question 1465 Marks
Differentiate the following with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
AnswerLet $\text{y}=\cot^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta,\text{ So}$
$\text{y}=\cot^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\cot^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg)$
$=\cot^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]$
$=\cot^{-1}\Big[\cot\Big(\frac{\pi}{2}-\frac{\pi}{4}+\theta\Big)\Big]$
$=\frac{\pi}{4}+\theta$
$\text{y}=\frac{\pi}{4}+\tan^{-1}\text{x}\ \big[\text{Since x}=\tan\theta\big]$
Differentiating it with respect do x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
View full question & answer→Question 1475 Marks
If the derivative of $\tan^{-1} (a + bx)$ takes the value 1 at x = 0, prove that $1 + a^2 = b.$
AnswerHere, $\frac{\text{d}}{\text{dx}}\big[\tan^{-1}(\text{a}+\text{bx})\big]=1\text{ at x}=0$
So, using chain rule,
$\Big[\Big\{\frac{1}{1+(\text{a}+\text{bx})^2}\Big\}\frac{\text{d}}{\text{dx}}(\text{a}+\text{bx})\Big]_{\text{x}=0}=0$
$\Big[\frac{1}{1+(\text{a}+\text{bx})^2}\times(\text{b})\Big]_{\text{x}=0}=1$
$\Rightarrow \frac{\text{b}}{1+(\text{a}+0)^2}=1$
$\Rightarrow \text{b}=1+\text{a}^2$
View full question & answer→Question 1485 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}$
AnswerHere,
$\text{y}=\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x})$
$\log\text{y}=\log\sin\text{x}+\log\sin2\text{x}+\log\sin3\text{x}+\log\sin4\text{x}$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\frac{\text{d}}{\text{dx}}\log\sin3\text{x}+\frac{\text{d}}{\text{dx}}\log\sin4\text{x}$
$=\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\frac{1}{\sin3\text{x}}\frac{\text{d}}{\text{dx}}(\sin3\text{x})+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4 \text{x})$
$=\frac{1}{\sin\text{x}}(\cos\text{x})+\frac{1}{\sin2\text{x}}(\cos2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x}) \\ +\frac{1}{\sin3\text{x}}(\cos3\text{x})\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\big[\cot\text{x}+\cot2\text{x}(2)+\cot3\text{x}(3)+\cot4\text{x}(4)\big]$
$\frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}) \\ \big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
[Using equation (i)]
View full question & answer→Question 1495 Marks
If $\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}$
AnswerHere,
$\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}}$
$\text{y}=(\cos\text{x})^\text{y}$
Taking log on both the sides,
$\log\text{y}=\log(\cos\text{x})^\text{y}$
$\log\text{y}=\text{y}\log(\cos\text{x}),\big\{\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big\}$
Differentiating it with resepect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\log(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{1}{\cos\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\cos\text{x}\Big)=\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\log\cos\text{x}}{\text{y}}\Big)=-\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{(1-\log\cos\text{x})}$
View full question & answer→Question 1505 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
AnswerLet $\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)$ To find the domain, we need to find all x such that $-1\leq\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ Since the quantity in the middle is always psitive, we need to find all x such that $\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ i.e. all x such that $2^{\text{x}+1}\leq1+4^\text{x}$ We may req. write as $2\leq\frac{1}{2^\text{x}}+2^\text{x},$ which is true for all x Hence, the function is defined at all real numbers. Putting $2^\text{x}=\tan\theta$$\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)=\sin^{-1}\Big(\frac{2^\text{x}.2}{1+(2^\text{x})^2}\Big)$
$=\sin^{-1}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}(2^\text{x})$ Thus, $\text{f(x)}=2\frac{1}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$ $=\frac{2}{1+4^\text{x}}(2^\text{x})\log2=\frac{2^{\text{x}+1}\log2}{1+4^\text{x}}$
View full question & answer→