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.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$e^x(y+1) = 1$
(a) $y$
(b) $y+1$
(c) $\dfrac{dy}{dx}$
(d) $\dfrac{dy}{dx}+1$
›Reveal solutionSolution

Solve for $y$ explicitly, differentiate twice, and express the result back in terms of $y$ .

$e^x(y+1)=1\Rightarrow y+1=e^{-x}\Rightarrow y=e^{-x}-1$ .

…

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GUJCET 2022Set 081 markMCQ
$y=100e^{2x}+200e^{-2x}$
›Reveal solutionSolution

Differentiating twice each exponential brings down a factor $4$ , giving $y''=4y$ .

Concept. $y=100e^{2x}+200e^{-2x}$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$e^y(x + 1) = 1$
(a) $-\left(\frac{dy}{dx}\right)$
(b) $\left(\frac{dy}{dx}\right)$
(c) $-\left(\frac{dy}{dx}\right)^2$
(d) $\left(\frac{dy}{dx}\right)^2$
›Reveal solutionSolution

Solve for $y$ explicitly, then relate $y''$ to $(y')^2$ .

$e^y(x+1) = 1 \Rightarrow e^y = \dfrac{1}{x+1} \Rightarrow y = -\ln(x+1).$

$\dfrac{dy}{dx} = -\dfrac{1}{x+1},\qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{(x+1)^2}.$

…

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GUJCET 2021Set 151 markMCQ
$x + 1 = e^{-y}$
›Reveal solutionSolution

Differentiate twice and recognize $\dfrac{d^2y}{dx^2}=\left(\dfrac{dy}{dx}\right)^2$ .

Concept: $x+1=e^{-y}\Rightarrow y=-\log(x+1)$ , so

$\frac{dy}{dx}=-\frac{1}{x+1},\qquad \frac{d^2y}{dx^2}=\frac{1}{(x+1)^2}.$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$y=\log_e(\log_e x)$
(a) $-\dfrac{\log_e(ex)}{(x\cdot\log_e x)^2}$
(b) $\dfrac{\log_e(ex)}{(x\cdot\log_e x)^2}$
(c) $-\dfrac{(x\cdot\log_e x)^2}{\log_e(ex)}$
(d) $\dfrac{\log_e(e/x)}{(x\cdot\log_x x)^2}$
›Reveal solutionSolution

Differentiate $y=\ln(\ln x)$ twice using the chain and quotient rules, then recognise $1+\ln x=\ln(ex)$ .

$y=\ln(\ln x)$ . First derivative: $y'=\dfrac{1}{\ln x}\cdot\dfrac1x=\dfrac{1}{x\ln x}$ .

Second derivative, using $y'=(x\ln x)^{-1}$ : $y''=-(x\ln x)^{-2}\cdot\dfrac{d}{dx}(x\ln x)$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$y = \tan^{-1} x$
(a) $-2x y_1$
(b) $x y_1$
(c) $2x y_1$
(d) $-x y_1$
›Reveal solutionSolution

Differentiate the first-order relation to get a relation for the second derivative.

For $y=\tan^{-1}x$ , $y_1=\dfrac{1}{1+x^2}$ , so $(1+x^2)y_1=1$ .

Differentiate both sides: …

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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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