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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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

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Sol. (1)

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Comments
  • March 14, 2021 at 8:31 pm

    and one more thing add some more questions……/////

    21
  • March 14, 2021 at 8:30 pm

    can you send me the solution of last question and 7th question…..

    29
  • February 23, 2021 at 5:04 pm

    these question answer have not full solution understand for an average student…….

    5
    • ...
      March 14, 2021 at 8:32 pm

      right

      0
  • December 13, 2020 at 9:52 am

    Good

    0
  • December 9, 2020 at 7:45 am

    Nice stuff amd content add some more questions

    4
    • February 23, 2021 at 5:04 pm

      not so much……

      1
  • September 28, 2020 at 4:51 pm

    some more questions

    2
  • September 22, 2020 at 9:41 am

    There are some typing mistakes I mean one letter is on another letter

    3
  • September 8, 2020 at 7:46 am

    Good one

    2
  • September 6, 2020 at 5:16 pm

    Nice thanks for uploding this quetions on Google

    1
  • September 4, 2020 at 10:59 pm

    bakvas

    0
  • July 28, 2020 at 11:35 am

    super

    1
  • July 15, 2020 at 11:49 am

    8th question was incorrect its f(1/2) please see it once

    1
  • June 27, 2020 at 3:48 pm

    Nice

    1