Differentiation JEE Main previous year questions with solutions

4 solved JEE Main questions on Differentiation, free to read — no sign-in needed. The full chapter has 73 questions; sign in to attempt the remaining 69 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Functional Equation
   Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbf{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
   1. A.$20$
   2. B.$40$
   3. C.$-20$
   4. D.$-40$
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   Answer: (B)

   Given $f''(x) = g''(x)$ for all $x \in \mathbf{R}$. Let $h(x) = f(x) - g(x)$. Taking the second derivative, $h''(x) = f''(x) - g''(x) = 0$. Integrating with respect to $x$, $h'(x) = c_1$. Given $f'(1) = 4$ and $g'(1) = 2$, $h'(1) = f'(1) - g'(1) = 4 - 2 = 2$. Therefore, $c_1 = 2$, which gives $h'(x) = 2$. Integrating again with respect to $x$, $h(x) = 2x + c_2$. Given $3f(2) = 9 \Rightarrow f(2) = 3$ and $g(2) = 9$, $h(2) = f(2) - g(2) = 3 - 9 = -6$. Substituting $x = 2$ in $h(x)$, $h(2) = 2(2) + c_2 = -6 \Rightarrow c_2 = -10$. Thus, $h(x) = 2x - 10$. Substituting $x = 25$, $h(25) = f(25) - g(25) = 2(25) - 10 = 40$. Answer: $40$
2. Q2JEE Main 2024 (09 Apr, Shift 2)Differentiation of composite functions
   If $\log _e y=3 \sin ^{-1} x$, then $\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}$ at $x=\frac{1}{2}$ is equal to
   1. A.$3 e^{\pi / 6}$
   2. B.$9 e^{\pi / 2}$
   3. C.$3 e^{\pi / 2}$
   4. D.$9 e^{\pi / 6}$
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   Answer: (B)

   $\begin{aligned} & \ln (y)=3 \sin ^{-1} x \\ & \frac{1}{y} \cdot y^{\prime}=3\left(\frac{1}{\sqrt{1-x^2}}\right) \\ & \Rightarrow y^{\prime}=\frac{3 y}{\sqrt{1-x^2}} \text { at } x=\frac{1}{2} \\ & \Rightarrow y^{\prime}=\frac{3 e^{3\left(\frac{\pi}{6}\right)}}{\frac{\sqrt{3}}{2}}=2 \sqrt{3} e^{\frac{\pi}{2}} \\ & \Rightarrow y^{\prime \prime}=3\left(\frac{\sqrt{1-x^2} y^{\prime}-y \frac{1}{2 \sqrt{1-x^2}}(-2 x)}{\left(1-x^2\right)}\right)\end{aligned}$ $\begin{aligned} & \Rightarrow\left(1-x^2\right) y^{\prime \prime}=3\left(3 y+\frac{x y}{\sqrt{1-x^2}}\right) \\ & \downarrow \text { at } x=\frac{1}{2}, y=e^{3 \sin ^{-1}\left(\frac{1}{2}\right)}=\mathrm{e}^{3\left(\frac{\pi}{6}\right)}=\mathrm{e}^{\frac{\pi}{2}}\end{aligned}$ $\begin{aligned} & \left.\left(1-x^2\right) y^{\prime \prime}\right|_{\text {at } x=\frac{1}{2}}=3\left(3 \mathrm{e}^{\frac{\pi}{2}}+\frac{\frac{1}{2}\left(e^{\frac{\pi}{2}}\right)}{\frac{\sqrt{3}}{2}}\right) \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right) \\ & \left(1-x^2\right) y^{\prime \prime}-\left.x y^{\prime}\right|_{\text {atx } x=\frac{1}{2}} \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\left(2 \sqrt{3} \mathrm{e}^{\frac{\pi}{2}}\right)=9 \mathrm{e}^{\frac{\pi}{2}}\end{aligned}$
3. Q3JEE Main 2022 (25 Jun, Shift 1)Derivative of function and its inverse
   Let $f:R\rightarrow R$ be defined as $f\left(x\right)={x}^{3}+x-5$. If $g\left(x\right)$ is a function such that $f\left(g\left(x\right)\right)=x,\forall x\in R$, then ${g}^{'}\left(63\right)$ is equal to ______
   1. A.$49$
   2. B.$\dfrac{1}{49}$
   3. C.$\dfrac{43}{49}$
   4. D.$\dfrac{3}{49}$
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   Answer: (B)

   Here, $g\left(f\left(x\right)\right)=x$ as $f\left(x\right)$ and $g\left(x\right)$ are inverse of each other. Now, ${g}^{'}\left(f\left(x\right)\right){f}^{'}\left(x\right)=1$ $\Rightarrow {g}^{'}\left(f\left(x\right)\right)=\dfrac{1}{{f}^{'}\left(x\right)}\ldots \left(i\right)$ Now $f\left(x\right)=63\Rightarrow {x}^{3}+x-5=63$ $\Rightarrow {x}^{3}+x-68=0$ So $x=4$ satisfies the above equation ${g}^{'}\left(63\right)=\dfrac{1}{{f}^{'}\left(4\right)}$ from $\left(i\right)$ $=\dfrac{1}{3{\left(4\right)}^{2}+1}=\dfrac{1}{49}$
4. Q4JEE Main 2021 (26 Aug, Shift 1)Differentiation of implicit functions
   If $y=y(x)$ is an implicit function of $x$ such that ${\log }_{e}(x+y)=4xy$, then $\dfrac{{d}^{2}y}{d{x}^{2}}$ at $x=0$ is equal to
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   Answer: 40

   Given: ${\log }_{e}(x+y)=4xy$ When $x=0$, then $y=1$ ${\log }_{e}\left(x+y\right)=4xy$ $\Rightarrow x+y={e}^{4xy}$ Now differentiate w.r.t. $x$ $1+{y}^{'}={e}^{4xy}\left(4y+4x{y}^{'}\right)\ldots \left(i\right)$ At $(0,1)\Rightarrow {y}^{'}(0)+1=4\Rightarrow {y}^{'}(0)=3$ Now, again differentiate equation $(i)$, we get ${y}^{"}={e}^{4xy}{\left(4y+4x{y}^{'}\right)}^{2}+{e}^{4xy}\left(4{y}^{'}+4{y}^{'}+4x{y}^{"}\right)$ At $\left(0,1\right)$ ${y}^{"}(0)=1(4\times 1+0{)}^{2}+1(4\times 3+4\times 3+0)$ $\Rightarrow {y}^{"}(0)=16+24=40$ $\Rightarrow {y}^{"}(0)=40$