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.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set A1 markMCQ
$\frac{d^2}{dx^2}(\sin 2x) =$
(a) $4\sin 2x$
(b) $4\cos^2 2x$
(c) $-4\sin 2x$
(d) $2\sin 4x$
›Reveal solutionSolution

$\frac{d^2}{dx^2}(\sin 2x) = -4\sin 2x$ .

First derivative (chain rule):

$\frac{d}{dx}(\sin 2x) = 2\cos 2x.$

Second derivative: …

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CBSE 2026Set ANNUAL1 mark
$Q. If y = 8e⁻³ˣ, find d²y/dx².$
›Reveal solutionSolution

Differentiate $y=8e^{-3x}$ twice using the chain rule.

$\dfrac{dy}{dx} = 8\cdot(-3)e^{-3x} = -24e^{-3x}$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the second derivative for the function y=sin x + e^{2x}.$
›Reveal solutionSolution
$y''=-\sin x+4e^{2x}$ .

Concept. The second derivative is found by differentiating the first derivative; use $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}e^{kx}=k\,e^{kx}$ .

Steps.

$y=\sin x+e^{2x}$ .

First derivative: $y'=\cos x+2e^{2x}$ . …
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CBSE 2025Set ANNUAL1 mark
$y = \log x$
›Reveal solutionSolution

Differentiate $y=\log x$ twice.

$y=\log x \implies \dfrac{dy}{dx}=\dfrac1x$

…

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CBSE 2025Set ANNUAL1 markMCQ
$y = 2\sin x + 3\cos x$
(a) $y$
(b) $\dfrac{1}{y}$
(c) $-y$
(d) $-\dfrac{1}{y}$
›Reveal solutionSolution

Differentiate twice — the second derivative comes back around to -y, a classic SHM-type result.

$y = 2\sin x + 3\cos x$

$\dfrac{dy}{dx} = 2\cos x - 3\sin x$

…

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CBSE 2025Set ANNUAL1 mark
$x\cos x$
›Reveal solutionSolution

Apply the product rule twice.

Let $y=x\cos x$ .

First derivative (product rule on $x$ and $\cos x$ ):

$y'=\frac{d}{dx}(x)\cos x+x\frac{d}{dx}(\cos x)=1\cdot\cos x+x(-\sin x)=\cos x-x\sin x.$

Second derivative (differentiate $\cos x$ and the product $x\sin x$ ): …

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CBSE 2024Set D1 markMCQ
$y = x^{20}$
(a) $x^{18}$
(b) $20x^{19}$
(c) $380x^{18}$
(d) $x^{19}$
›Reveal solutionSolution

$\frac{d^2y}{dx^2} = 380x^{18}$ .

Use the power rule $\frac{d}{dx}x^n = nx^{n-1}$ twice.

First derivative:

$\frac{dy}{dx} = 20x^{19}.$

Second derivative: …

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CBSE 2024Set ANNUAL1 markMCQ
$y=\log x$
(a) $-\frac{1}{x^2}$
(b) $\frac{1}{x^2}$
(c) $-\frac{1}{x}$
(d) $\frac{1}{x}$
›Reveal solutionSolution

Differentiate y=log x twice: first derivative is 1/x, second derivative is -1/x^2.

$y=\log x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x}=x^{-1}$

…

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CBSE 2024Set ANNUAL1 mark
$\log x$
›Reveal solutionSolution

Differentiate $\log x$ twice.

Let $y = \log x$ . The first derivative is

$\dfrac{dy}{dx} = \dfrac{1}{x} = x^{-1}.$

Differentiating again, …

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CBSE 2023Set E1 markMCQ
$\dfrac{d^{2}}{dx^{2}}\left(e^{5x}\right) =$
(a) $e^{5x}$
(b) $10e^{5x}$
(c) $5e^{5x}$
(d) $25e^{5x}$
›Reveal solutionSolution

$\dfrac{d}{dx}e^{5x}=5e^{5x}$ , and differentiating again gives $25e^{5x}$ .

First derivative: $\dfrac{d}{dx}e^{5x}=5e^{5x}$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$y=x \cdot \log_e x$
(a) $\frac{1}{1+x}$
(b) $\frac{1}{x}$
(c) $\log_e(1+x)$
(d) $1+\log_e x$
›Reveal solutionSolution

Differentiate $y=x\log_e x$ twice using the product rule.

Given $y=x\log_e x$ .

First derivative (product rule, $u=x$ , $v=\log_e x$ ):

$\dfrac{dy}{dx}=1\cdot\log_e x + x\cdot\dfrac{1}{x}=\log_e x+1$

…

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CBSE 2022Set ANNUAL1 markMCQ
$\dfrac{d^2}{dx^2}\sin(3x+5) = ?$
(a) $\sin(3x+5)$
(b) $9\cos(3x+5)$
(c) $-9\sin(3x+5)$
(d) $9\sin(3x+5)$
›Reveal solutionSolution

Differentiate twice by the chain rule; each derivative brings down a factor of 3, and differentiating sine twice turns it into $-\sin$ .

Let $y=\sin(3x+5)$ .

First derivative: $\dfrac{dy}{dx}=3\cos(3x+5)$ .

…

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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

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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