Method Of Differentiation – JEE Main Previous Year Question with Solutions

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Q. Let y be an implicit function of x defined by $x^{2 x}-2 x^{x}$ cot $y-1=0 .$ Then $y^{\prime}(1)$ equals :

(1) $\log 2$

$(2)-\log 2$

(3) –1

(4) 1

[AIEEE-2009]

Sol. (3)

$\left(x^{x}\right)^{2}-2 \cot y x^{x}-1=0$

$\mathrm{x}^{\mathrm{x}}=\frac{2 \cot y \pm \sqrt{4 \cot ^{2} y+4}}{2}\left\{\begin{array}{l}{\text { at } \mathrm{x}=1} \\ {1=\cot y+\csc y} \\ {\Rightarrow y=\frac{\pi}{2}}\end{array}\right.$

$=\cot y \pm \csc y$

$\mathrm{x}^{\mathrm{x}}=\cot \mathrm{y}+\csc \mathrm{y}$

diff. w.r. to $\mathrm{x}$

$\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})=\left[-\csc ^{2} \mathrm{y}-\csc \mathrm{y} \cot \mathrm{y}\right] \frac{\mathrm{d} y}{\mathrm{dx}}$

$1=-\csc y[\csc y+\cot y] \frac{d y}{d x}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=-1$

Q. Let $\mathrm{f}:(-1,1) \rightarrow \mathrm{R}$ be a differentiable function with $\mathrm{f}(0)=-1$ and $\mathrm{f}(0)=1 .$ Let $\mathrm{g}(\mathrm{x})=[\mathrm{f}(2 \mathrm{f}(\mathrm{x})+$ $2)]^{2} .$ Then $\mathrm{g}^{\prime}(0):$

(1) 4            (2) –4             (3) 0              (4) –2

[AIEEE-2010]

Sol. (2)

$\mathrm{g}(\mathrm{x})=[\mathrm{f}(2 \mathrm{f}(\mathrm{x})+2)]^{2}$

$\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{f}[2 \mathrm{f}(\mathrm{x})+2] \mathrm{f}^{\prime}[2 \mathrm{f}(\mathrm{x})+2] \cdot 2 \mathrm{f}^{\prime}(\mathrm{x})$

Put $\mathrm{x}=0$

$\mathrm{g}^{\prime}(0)=2 \mathrm{f}[2 \mathrm{f}(0)+2] \mathrm{f}^{\prime}[2 \mathrm{f}(0)+2] 2 \mathrm{f}^{\prime}(0)$

$=2 \mathrm{f}(2(-1)+2) \mathrm{f}^{\prime}(2(-1)+2) 2 \mathrm{f}^{\prime}(0)$

$=2 \mathrm{f}(0) \mathrm{f}^{\prime}(0) 2 \mathrm{f}^{\prime}(0)$

$=4(-1)(1)(1)=-4$

Q. $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}$ equals :

(1) $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2}$

(2) $-\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$

(3) $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{-1}$

$(4)-\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$

[AIEEE-2011]

Sol. (2)

$\frac{\mathrm{d}}{\mathrm{dy}}\left(\left(\frac{\mathrm{d} y}{\mathrm{dx}}\right)^{-1}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\left(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-1}\right) \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$

$=-\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2} \cdot \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dx}^{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}=-\left(\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$

Q. If $\mathrm{y}=\sec \left(\tan ^{-1} \mathrm{x}\right),$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=1$ is equal to :

(1) $\frac{1}{\sqrt{2}}$

(2) $\frac{1}{2}$

(3) 1

(4) $\sqrt{2}$

[JEE-MAIN-2013]

Sol. (1)

$\mathrm{y}=\sec \left(\tan ^{-1} \mathrm{x}\right)=\sqrt{1+\mathrm{x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$

$\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=\frac{1}{\sqrt{2}}$

Q. If $\mathrm{g}$ is the inverse of a function $f$ and $f^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{5}},$ then $\mathrm{g}^{\prime}(\mathrm{x})$ is equal to :

(1) $1+x^{5}$

(2) $5 x^{4}$

(3) $\frac{1}{1+\{\mathrm{g}(\mathrm{x})\}^{5}}$

(4) $1+\{\mathrm{g}(\mathrm{x})\}^{5}$

[JEE-MAIN-2014]

Sol. (4)

$\mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{x}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \cdot \mathrm{g}^{\prime}(\mathrm{x})=1$

$\therefore \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x}))}=1+\{\mathrm{g}(\mathrm{x})\}^{5}$

Q. If for $x \in\left(0, \frac{1}{4}\right),$ the derivative of $\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right)$ is $\sqrt{x} \cdot g(x),$ then $g(x)$ equals :

(1) $\frac{3}{1+9 x^{3}}$

(2) $\frac{9}{1+9 x^{3}}$

(3) $\frac{3 x \sqrt{x}}{1-9 x^{3}}$

(4) $\frac{3 \mathrm{x}}{1-9 \mathrm{x}^{3}}$

[JEE-MAIN-2017]

Sol. (2)

Let $\mathrm{y}=\tan ^{-1}\left(\frac{6 \mathrm{x} \sqrt{\mathrm{x}}}{1-9 \mathrm{x}^{3}}\right)$ where $\mathrm{x} \in\left(0, \frac{1}{4}\right)$

$=\tan ^{-1}\left(\frac{2 \cdot\left(3 \mathrm{x}^{3 / 2}\right)}{1-\left(3 \mathrm{x}^{3 / 2}\right)}\right)=2 \tan ^{-1}\left(3 \mathrm{x}^{3 / 2}\right)$

As $3 x^{3 / 2} \in\left(0, \frac{3}{8}\right)$

$\therefore \frac{d y}{d x}=2 \times \frac{1}{1+9 x^{3}} \times 3 \times \frac{3}{2} \times x^{1 / 2}$

\begin{aligned} &=\frac{9}{1+9 \mathrm{x}^{3}} \sqrt{\mathrm{x}} \\ \therefore \mathrm{g}(\mathrm{x}) &=\frac{9}{1+9 \mathrm{x}^{3}} \end{aligned}

Q. If $\mathrm{x}=\sqrt{2^{\cos \mathrm{c}^{-1} \mathrm{t}}}$ and $\mathrm{y}=\sqrt{2^{\mathrm{sc}-\mathrm{t}}}(|\mathrm{t}| \geq 1),$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to :

(1) $-\frac{y}{x}$

( 2)$\frac{\mathrm{x}}{\mathrm{y}}$

$(3)-\frac{x}{y}$

(4) $\frac{y}{x}$

[JEE-MAIN-2018]

Sol. (1)

Q. If $\mathrm{f}(\mathrm{x})=\sin ^{-1}\left(\frac{2 \times 3^{\mathrm{x}}}{1+9^{\mathrm{x}}}\right),$ then $\mathrm{f}^{\prime}\left(-\frac{1}{2}\right)$ equals :

(1) $\sqrt{3} \log _{e} \sqrt{3}$

(2) $-\sqrt{3} \log _{e} 3$

(3) $-\sqrt{3} \log _{e} \sqrt{3}$

(4) $\sqrt{3} \log _{e} 3$

[JEE-MAIN-2018]

Sol. (1)

Q. If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{cc}{\cos \mathrm{x}} & {\mathrm{x}} \\ {2 \sin \mathrm{x}} & {\mathrm{x}^{2} 2 \mathrm{x}} \\ {\tan \mathrm{x}} & {\mathrm{x}} & {1}\end{array}\right|,$ then $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{x}}$

(1) exists and is equal to 0

(2) exists and is equal to –2

(3) exists and is equal to 2

(4) does not exist

[JEE-MAIN-2018]

Sol. (2)

Q. If $x^{2}+y^{2}+\sin y=4,$ then the value of $\frac{d^{2} y}{d x^{2}}$ at the point $(-2,0)$ is

(1) –34             (2) –32            (3) –2               (4) 4

Sol. (1)