Method Of Differentiation – JEE Main Previous Year Question with Solutions
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Administrator

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