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Method Of Differentiation - JEE Main Previous Year Question with Solutions

Method of Differentiation covers essential JEE Main concepts such as implicit differentiation, inverse functions, chain rule, parametric differentiation, higher-order derivatives, and applications through previous year solved questions.

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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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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
Ans. (1)

Frequently Asked Questions

Find answers to common questions.

How many questions come from Method of Differentiation in JEE Main?

NTA typically sets 1–2 questions directly from Method of Differentiation per session, carrying 4–8 marks. Indirect applications in Tangents and Normals, Rate of Change, and Maxima–Minima can bring the effective contribution to 10–12 marks. Consistent practice with PYQs from 2009 onwards gives the best preparation for this range.

What are the most important differentiation techniques for JEE Main?

The five most tested techniques are: (1) chain rule with inverse trigonometric functions, (2) implicit differentiation, (3) parametric differentiation, (4) logarithmic differentiation for expressions like x^x, and (5) differentiation of determinants. Questions on second-order derivatives frequently combine two of these techniques in a single problem.

How do I differentiate implicit functions for JEE Main?

Differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule wherever y appears. Then collect all dy/dx terms on one side and solve. Always substitute the given point back into the original equation first to find the exact value of y before computing the numerical answer.

Is Method of Differentiation in JEE Main syllabus 2025?

Yes. Method of Differentiation is explicitly listed under Calculus in the NTA JEE Main 2025 official syllabus. It falls under Unit 10 — Differential Calculus — and includes derivatives of all standard functions, rules of differentiation, higher-order derivatives, and implicit and parametric differentiation.

What is the trick for inverse function derivatives in JEE Main?

If g is the inverse of f, use the identity g′(x) = 1/f′(g(x)). This avoids computing g explicitly. In Q5 above, since f′(x) = 1/(1+x⁵), applying this identity directly gives g′(x) = 1 + {g(x)}⁵ in two lines — no complex algebra needed.

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Comments

Sarthak
Feb. 13, 2025, 11:22 a.m.
Acche question give me
mufassira
Aug. 26, 2021, 8:41 p.m.
sir some of these quetsions solutions have not given can you please make sure that solutions should contain
techno
March 14, 2021, 8:31 p.m.
and one more thing add some more questions....../////
techno
March 14, 2021, 8:30 p.m.
can you send me the solution of last question and 7th question.....
lucky gaur
Feb. 23, 2021, 5:04 p.m.
these question answer have not full solution understand for an average student.......
Csct
Dec. 13, 2020, 9:52 a.m.
Good
Allu arjun
Dec. 9, 2020, 7:45 a.m.
Nice stuff amd content add some more questions
NIHARIKA
Sept. 28, 2020, 4:51 p.m.
some more questions
Saketh
Sept. 22, 2020, 9:41 a.m.
There are some typing mistakes I mean one letter is on another letter
Mahesh
Sept. 8, 2020, 7:46 a.m.
Good one
Sakshi
Sept. 6, 2020, 5:16 p.m.
Nice thanks for uploding this quetions on Google
navyug
Sept. 4, 2020, 10:59 p.m.
bakvas
ravi
July 28, 2020, 11:35 a.m.
super
Alekhya
July 15, 2020, 11:49 a.m.
8th question was incorrect its f(1/2) please see it once
Akash
June 27, 2020, 3:48 p.m.
Nice
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