If $y = \tan^{-1} x$, find $\dfrac{d^2 y}{dx^2}$ in terms of $y$ alone.

If y = tan⁻¹ x, find (d² y)/(dx²) in terms of y alone.

NCERT Exemplar · Q64

y = \tan^{-1} x

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Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

Differentiate a function, then differentiate the result, then differentiate that, and so on. Each pass produces a new function describing a deeper layer of change. The physics picture makes it concrete: position $\to$ velocity (1st derivative) $\to$ acceleration (2nd) $\to$ jerk (3rd). Every step asks the same question: "how does the previous rate of change itself change?"

The definition and notation

If $y=f(x)$ , its successive derivatives are written $y_1,y_2,\dots,y_n$ , equivalently $f'(x),f''(x),\dots,f^{(n)}(x)$ or $\dfrac{dy}{dx},\dfrac{d^2y}{dx^2},\dots,\dfrac{d^ny}{dx^n}$ . Each one is the derivative of the previous:

\frac{d^ny}{dx^n}=\frac{d}{dx}\!\left(\frac{d^{n-1}y}{dx^{n-1}}\right).

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Seeing the pattern

Take $y=x^4$ : $\;y_1=4x^3,\;y_2=12x^2,\;y_3=24x,\;y_4=24,\;y_5=0$ . Each differentiation drops the degree by one, so a degree- $n$ polynomial has a constant $n$ th derivative and vanishing higher ones. Other families behave differently: $e^{ax}$ never dies out ( $\frac{d^n}{dx^n}e^{ax}=a^n e^{ax}$ ), and $\sin x$ cycles every four steps ( $\sin\to\cos\to-\sin\to-\cos$ ).

Tip

Power rule applied $n$ times: $\dfrac{d^n}{dx^n}(x^m)=m(m-1)\cdots(m-n+1)\,x^{m-n}$ for $n\le m$ .

Watch out

$\dfrac{d^2y}{dx^2}$ is not $\left(\dfrac{dy}{dx}\right)^2$ — a second derivative is not the square of the first derivative. …

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Writing $x=\tan y$ gives $\dfrac{dy}{dx}=\cos^2 y$ ; differentiating again gives $\dfrac{d^2y}{dx^2}=-2\sin y\cos^3 y=-\sin 2y\,\cos^2 y$ .

First derivative. If $y=\tan^{-1}x$ then $x=\tan y$ . Differentiating $x=\tan y$ with respect to $x$ :

1=\sec^2 y\,\frac{dy}{dx}\quad\Longrightarrow\quad \frac{dy}{dx}=\frac{1}{\sec^2 y}=\cos^2 y.

Second derivative. Differentiate $\dfrac{dy}{dx}=\cos^2 y$ with respect to $x$ , remembering $y$ is a function of $x$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04174 marksMCQ
$y = 4e^{-x} - 2e^{-2x} - e^{-3x}$
›Reveal solutionSolution

Differentiate twice and factor out $e^{-3x}$ ; the exponent is $\alpha=-3$ .

With $y=4e^{-x}-2e^{-2x}-e^{-3x}$ :
$\frac{dy}{dx}=-4e^{-x}+4e^{-2x}+3e^{-3x},$ $\frac{d^2y}{dx^2}=4e^{-x}-8e^{-2x}-9e^{-3x}.$
Factor the common $e^{-3x}$ : …
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KEAM 2026Set eng-2026-04174 marksMCQ
$y = 4\sqrt{x}$
›Reveal solutionSolution

Compute the two derivatives, use $y^2=16x$ to eliminate $x$ .

With $y=4x^{1/2}$ :
$\frac{dy}{dx}=2x^{-1/2},\qquad \frac{d^2y}{dx^2}=-x^{-3/2}.$
Also $y^2=16x\Rightarrow \dfrac{1}{y^2}=\dfrac{1}{16x}$ . Then

$\frac{1}{y^2}\frac{dy}{dx}=\frac{2x^{-1/2}}{16x}=\frac{1}{8}x^{-3/2}.$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$y = e^{-x^2}$
›Reveal solutionSolution

Differentiate $y=e^{-x^2}$ twice, keeping results in terms of $y$ ; the combination $y''+2xy'$ collapses to $-2y$ .

Concept. Because $y=e^{-x^2}$ satisfies $y'=-2xy$ , its higher derivatives can be written back in terms of $y$ itself — a differential-equation viewpoint that makes the given combination easy to evaluate.

Step 1 — first derivative (chain rule).
$\frac{dy}{dx}=e^{-x^2}\cdot(-2x)=-2x\,e^{-x^2}=-2xy.$
Step 2 — second derivative (product rule on $-2xy$ ). …
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KEAM 2026Set eng-2026-04204 marksMCQ
$g(x) = \begin{vmatrix} x^4 & -\cos x & -\sin 2x \\ -8 & 4 & 3 \\ 2 & 4 & 8 \end{vmatrix}$
›Reveal solutionSolution

Expand the determinant along the top row, then differentiate three times and set $x=0$ .

Cofactors of row 1: $\det\begin{pmatrix}4&3\\4&8\end{pmatrix}=20$ , $-\det\begin{pmatrix}-8&3\\2&8\end{pmatrix}=-(-70)=70$ , $\det\begin{pmatrix}-8&4\\2&4\end{pmatrix}=-40$ .

$g(x)=x^4(20)+(-\cos x)(70)+(-\sin 2x)(-40)=20x^4-70\cos x+40\sin 2x$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
$y = \sin x + e^x$
›Reveal solutionSolution

Invert the derivative and differentiate again w.r.t. $y$ via the chain rule: $\dfrac{d^2x}{dy^2}=\dfrac{\sin x-e^x}{(\cos x+e^x)^3}$ .

$\dfrac{dy}{dx}=\cos x+e^x\Rightarrow \dfrac{dx}{dy}=\dfrac{1}{\cos x+e^x}$ .

$\dfrac{d^2x}{dy^2}=\dfrac{d}{dy}\!\left(\dfrac{dx}{dy}\right)=\dfrac{d}{dx}\!\left(\dfrac{1}{\cos x+e^x}\right)\cdot\dfrac{dx}{dy}$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$f(x) = (2x + 3)e^{5x}$
›Reveal solutionSolution

Differentiate the product $f(x)=(2x+3)e^{5x}$ twice, evaluate at $x=1$ , and combine.

Given $f(x)=(2x+3)e^{5x}$ .

First derivative: $f'(x)=2e^{5x}+5(2x+3)e^{5x}=e^{5x}(10x+17)$ .

Second derivative: $f''(x)=5e^{5x}(10x+17)+10e^{5x}=e^{5x}(50x+95)$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$f(x)=\dfrac{1}{x^2}$
›Reveal solutionSolution

Compute $f''$ , form the product $u=f\,f''$ , then differentiate.

With $f(x)=x^{-2}$ :
$f'(x)=-2x^{-3},\qquad f''(x)=6x^{-4}.$
Then
$u=f(x)f''(x)=x^{-2}\cdot 6x^{-4}=6x^{-6},$
and …
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KEAM 2025Set eng-2025-04254 marksMCQ
$y=(x-1)\log_e(x-1)$
›Reveal solutionSolution

Differentiate the product twice; the second derivative is $\dfrac{1}{x-1}$ , giving $\dfrac{1}{2}$ at $x=3$ .

With $y=(x-1)\log_e(x-1)$ ,
$y'=\log_e(x-1)+(x-1)\cdot\dfrac{1}{x-1}=\log_e(x-1)+1.$
Differentiating again, …
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KEAM 2025Set eng-2025-04264 marksMCQ
$f: \mathbb{R} \to \mathbb{R}$
›Reveal solutionSolution

The cubic's third derivative is constant $6$ , so $f'''(3)=6$ .

Write $a=f'(1),\;b=f''(2),\;c=f'''(3)$ (all constants), so $f(x)=x^3+ax^2+bx+c$ .

Then $f'''(x)=6$ for all $x$ , hence $f'''(3)=6$ , i.e. $c=6$ . …
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KEAM 2025Set eng-2025-04284 marksMCQ
$y = (5x - 2)e^x$
›Reveal solutionSolution

Differentiate twice by the product rule: $y'=e^x(5x+3)$ , $y''=e^x(5x+8)$ .

Given $y=(5x-2)e^x$ . First derivative:
$y'=5e^x+(5x-2)e^x=e^x(5x-2+5)=e^x(5x+3).$
Second derivative: …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$\frac{d}{dx}\left(\frac{1}{x}\frac{d^2}{dx^2}\left(\frac{1}{x^3}\right)\right) =$
›Reveal solutionSolution
The expression equals $-72x^{-7}$ .

Concept and Intuition

Apply the power rule $\frac{d}{dx}x^n=nx^{n-1}$ repeatedly, working from the inside out.

Step-by-Step Solution

$\frac{d}{dx}x^{-3}=-3x^{-4}$ , then $\frac{d^2}{dx^2}x^{-3}=\frac{d}{dx}(-3x^{-4})=12x^{-5}$ .

Multiply by $\frac1x$ : $\frac1x\cdot12x^{-5}=12x^{-6}$ .

Differentiate: $\frac{d}{dx}(12x^{-6})=12(-6)x^{-7}=-72x^{-7}$ . …
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Q.If $y = \tan^{-1} x$ , find $\dfrac{d^2 y}{dx^2}$ in terms of $y$ alone.

Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

The definition and notation

Seeing the pattern

Method: Second Derivative of an Inverse Trigonometric Function via the Inverse Relation

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Q.If y=tan⁡−1xy = \tan^{-1} xy=tan−1x, find d2ydx2\dfrac{d^2 y}{dx^2}dx2d2y​ in terms of yyy alone.

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Q.If $y = \tan^{-1} x$ , find $\dfrac{d^2 y}{dx^2}$ in terms of $y$ alone.

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