Differentiation JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Differentiation, free to read — no sign-in needed. The full chapter has 10 questions; sign in to attempt the remaining 5 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $\mathbb{R}$ denote the set of all real numbers. Consider the polynomial function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \dfrac{d^{10}}{dx^{10}}\left((x^2 - 1)^{10}\right)$, for all $x \in \mathbb{R}$. Here $\dfrac{d^{10}}{dx^{10}}\left((x^2 - 1)^{10}\right)$ is the 10th order derivative of the function $(x^2 - 1)^{10}$. Then which of the following statements is (are) TRUE ?
   1. A.The coefficient of $x^8$ in the polynomial $f(x)$ is $(-10)\left(\dfrac{18!}{8!}\right)$
   2. B.The value of $f(1) + f(-1)$ is equal to $10!\, 2^{11}$
   3. C.The degree of the polynomial $f(x)$ is $10$
   4. D.The constant term of the polynomial $f(x)$ is $-\left(\dfrac{10!}{5!}\right)$
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   Answer: A,B,C

   Given $f(x) = \dfrac{d^{10}}{dx^{10}}\left((x^2 - 1)^{10}\right)$. Using the binomial expansion, we have: $(x^2 - 1)^{10} = \sum_{k=0}^{10} (-1)^k \, ^{10}C_{k} x^{20-2k}$ Differentiating $10$ times with respect to $x$, we get: $f(x) = \sum_{k=0}^{5} (-1)^k \, ^{10}C_{k} \dfrac{(20-2k)!}{(10-2k)!} x^{10-2k}$ For the coefficient of $x^8$, we set $10 - 2k = 8 \Rightarrow k = 1$. The coefficient is $(-1)^1 \, ^{10}C_{1} \dfrac{18!}{8!} = -10 \left(\dfrac{18!}{8!}\right)$. Thus, statement (A) is true. The highest power of $x$ in $f(x)$ corresponds to $k = 0$, which gives $x^{10}$ with a non-zero coefficient of $\dfrac{20!}{10!}$. Therefore, the degree of the polynomial $f(x)$ is $10$. Thus, statement (C) is true. For the constant term, we set $10 - 2k = 0 \Rightarrow k = 5$. The constant term is $(-1)^5 \, ^{10}C_{5} \dfrac{10!}{0!} = -\dfrac{10!}{5!5!} 10! = -\left(\dfrac{10!}{5!}\right)^2$. Thus, statement (D) is false. To find $f(1)$ and $f(-1)$, we use the Leibniz rule for the $n$-th derivative of a product: $f(x) = \dfrac{d^{10}}{dx^{10}}\left((x-1)^{10}(x+1)^{10}\right) = \sum_{r=0}^{10} \, ^{10}C_{r} \dfrac{d^{10-r}}{dx^{10-r}}\left((x-1)^{10}\right) \dfrac{d^r}{dx^r}\left((x+1)^{10}\right)$ Evaluating at $x = 1$, all terms in the sum are zero except when $r = 0$ (since $\dfrac{d^{10-r}}{dx^{10-r}}\left((x-1)^{10}\right)$ contains a factor of $(x-1)$ for $r \gt 0$). $f(1) = \, ^{10}C_{0} (10!) (1+1)^{10} = 10! \, 2^{10}$ Evaluating at $x = -1$, all terms are zero except when $r = 10$. $f(-1) = \, ^{10}C_{10} (-1-1)^{10} (10!) = 10! \, 2^{10}$ Adding these values gives: $f(1) + f(-1) = 10! \, 2^{10} + 10! \, 2^{10} = 10! \, 2^{11}$ Thus, statement (B) is true. Answer: The coefficient of $x^8$ in the polynomial $f(x)$ is $(-10)\left(\dfrac{18!}{8!}\right)$; The value of $f(1) + f(-1)$ is equal to $10!\, 2^{11}$; The degree of the polynomial $f(x)$ is $10$
2. Q2JEE Advanced Adv 2025 (Paper 2)
   Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow(0,4)$ be functions defined by $f(x)=\log _e\left(x^2+2 x+4\right)$, and $g(x)=\frac{4}{1+e^{-2 x}}$ Define the composite function $f \circ g^{-1}$ by $\left(f \circ g^{-1}\right)(x)=\left(g^{-1}(x)\right)$, where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $f \circ g^{-1}$ at $x=2$ is ________ .
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   Answer: 0.25

   $\begin{aligned} & \text { Let } \mathrm{h}(\mathrm{x})=\mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \text { and } \mathrm{g}(0)=2 \\ & \begin{aligned} \mathrm{h}^{\prime}(\mathrm{x}) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \cdot\left(\mathrm{g}^{-1}(\mathrm{x})\right)^{\prime} \\ \mathrm{h}^{\prime}(2) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(2) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\right. \\ & =\mathrm{f}^{\prime}(0) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\end{aligned}\end{aligned}$ Now $\begin{aligned} & f(x)=\log _0\left(x^2+2 x+4\right) \\ & f^{\prime}(x)=\frac{2 x+2}{x^2+2 x+4} \\ & f^{\prime}(0)=\frac{1}{2} \\ & g(x)=\frac{4}{1+e^{-2 x}}, g(0)=2 \\ & g^{-1}(g(x))=x \end{aligned}$ $$\begin{array}{l|l}\left(\left(\mathrm{g}^{-1}\right)^{\prime}(\mathrm{g}(\mathrm{x}))\right) \mathrm{g}^{\prime}(\mathrm{x})=1 & \mathrm{~g}(\mathrm{x})=\frac{4}{1+\mathrm{e}^{-2 \mathrm{x}}} \\\left(\mathrm{g}^{-1}\right)^{\prime}(2)=\frac{1}{\mathrm{~g}^{\prime}(0)} & \mathrm{g}^{\prime}(\mathrm{x})=\frac{8 \mathrm{e}^{-2 \mathrm{x}}}{\left(1+\mathrm{e}^{-2 \mathrm{x}}\right)^2} \\=\frac{1}{2} & \mathrm{~g}^{\prime}(0)=\frac{8}{4}=2 \\So \mathrm{h}(2)=\frac{1}{4}=0.25 &\end{array}$$
3. Q3JEE Advanced Adv 2016 (Paper 1)
   Let $f:R\rightarrow R,g:R\rightarrow R$ and $h:R\rightarrow R$ be differentiable functions such that $f\left(x\right)={x}^{3}+3x+2,g\left(f\left(x\right)\right)=x$ and $h\left(g\left(g\left(x\right)\right)\right)=x$ , for all $x\in R.$ Then,
   1. A.${g}^{'}\left(2\right)=\dfrac{1}{15}$
   2. B.${h}^{'}\left(1\right)=666$
   3. C.$h\left(0\right)=16$
   4. D.$h\left(g\left(3\right)\right)=36$
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   Answer: B,C

   If ${f}^{'}\left(x\right)=3{x}^{2}+3$ $\Rightarrow g\left(f\left(x\right)\right)=x$ $\Rightarrow {g}^{'}\left(f\left(x\right)\right)=\dfrac{1}{{f}^{'}\left(x\right)}$ $Also f\left(0\right)=2$ ($Put x=0$) So, ${g}^{'}\left(2\right)=\dfrac{1}{{f}^{'}\left(0\right)}$ $\Rightarrow$ $g'\left(2\right)=\dfrac{1}{3}$ If $h(g\left(g\left(x\right)\right)=x$ ${h}^{'}\left(g\left(g\left(x\right)\right)\right)=\dfrac{1}{{g}^{'}\left(g\left(x\right)\right).{g}^{'}\left(x\right)}$ If, $g\left(g\left(x\right)\right)=1$ $g(f\left(x\right))=x\Rightarrow {g}^{-1}\left(x\right)=f(x)$ $\Rightarrow g\left(x\right)={g}^{-1}\left(1\right)\Rightarrow g\left(x\right)=f\left(1\right)=6$ To find, ${g}^{'}\left(6\right)$ use $x=1$ in ${g}^{'}\left(f\left(x\right)\right)=\dfrac{1}{{f}^{'}\left(1\right)}$ as $f\left(1\right)=6$ To find, ${g}^{'}\left(236\right)$, use $x=6$ as $f\left(6\right)=236$ $∴{h}^{'}\left(1\right)=\dfrac{1}{{g}^{'}\left(6\right).{g}^{'}(236)}=\dfrac{1}{\dfrac{1}{6}.\dfrac{1}{111}}\Rightarrow {h}^{'}\left(1\right)=666$ If $g\left(g\left(x\right)\right)=0$ $∴g\left(x\right)={g}^{-1}\left(0\right)\Rightarrow g\left(x\right)=f\left(0\right)$ $\Rightarrow g\left(x\right)=2\Rightarrow x={g}^{-1}\left(2\right)$ $\Rightarrow x=f\left(2\right)\Rightarrow x=16$ $∴h\left(0\right)=16$ $g\left(x\right)=3\Rightarrow x={g}^{-1}\left(3\right)$ $\Rightarrow x=f\left(3\right)\Rightarrow x=38$ So $h\left(g\left(3\right)\right)=38$
4. Q4JEE Advanced Adv 2010 (Paper 2)
   Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad$ for all $x \in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to
   1. A.1
   2. B.$\frac{1}{3}$
   3. C.$\frac{1}{2}$
   4. D.$\frac{1}{e}$
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   Answer: (B)

   We have, $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t x \in(-1,1)$ On differentiating w.r.t. $x$, we get $$\begin{array}{ll} & e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\ \Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\ \because & f^{-1} \text { is the inverse of } f \\ \therefore & f^{-1}(f(x))=x \\ \Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\ \Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\ \Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\ \text { At } \quad & x=0, f(x)=2 \\ & f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3} \end{array}$$
5. Q5JEE Advanced Adv 2009 (Paper 2)
   If the function $f(x)=x^3+e^{\frac{x}{2}}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime}(1)$ is
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   Answer: 2

   Given, $g\{f(x)\}=x$ $$\begin{array}{lc} \Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\ \text { If } & f(x)=1 \Rightarrow x=0, f(0)=1 \end{array}$$ Substitute $x=0$ in Eq. (i), we get $$\begin{aligned} & g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\ & \Rightarrow \quad g^{\prime}(1)=2 \\ & {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\ & \end{aligned}$$