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$$(2)-\log 2(3) –1(4) 1 [AIEEE-2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\left(x^{x}\right)^{2}-2 \cot y x^{x}-1=0$$\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]

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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)

• August 26, 2021 at 8:41 pm

sir some of these quetsions solutions have not given
can you please make sure that solutions should contain

• March 14, 2021 at 8:31 pm

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

• March 14, 2021 at 8:30 pm

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

• February 23, 2021 at 5:04 pm

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

• March 14, 2021 at 8:32 pm

right

• December 13, 2020 at 9:52 am

Good

• December 9, 2020 at 7:45 am

Nice stuff amd content add some more questions

• February 23, 2021 at 5:04 pm

not so much……

• September 28, 2020 at 4:51 pm

some more questions

• September 22, 2020 at 9:41 am

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

• September 8, 2020 at 7:46 am

Good one

• September 6, 2020 at 5:16 pm

Nice thanks for uploding this quetions on Google

• September 4, 2020 at 10:59 pm

bakvas

• March 21, 2022 at 3:29 pm

Ok

• July 28, 2020 at 11:35 am

super

• July 15, 2020 at 11:49 am

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

• June 27, 2020 at 3:48 pm

Nice