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

Concept: Second derivative of inverse tangent expressed in terms of $y$.

We have $y = \tan^{-1} x$, so $x = \tan y$.

Step 1: Differentiate $x = \tan y$ with respect to $y$:

Step 2: Hence,

Step 3: Differentiate again with respect to $x$:

$\frac{d^2 y}{dx^2} = \frac{d}{dx}(\cos^2 y) = 2\cos y \cdot (-\sin y) \cdot \frac{dy}{dx}$ …

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

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

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

Use this method whenever $y$ is defined as an inverse trig function of $x$ (e.g. $y=\tan^{-1}x$, $y=\sin^{-1}x$) and you must find $\dfrac{d^2y}{dx^2}$ expressed in terms of $y$ itself, not $x$.

Steps

Step 1: Rewrite the inverse relation as $x$ in terms of $y$

If $y=\tan^{-1}x$, rewrite it as $x=\tan y$. Differentiating a standard trig function is easier than differentiating its inverse directly, so this flips the problem into an easier direction.

Step 2: Differentiate $x$ with respect to $y$, then invert

This uses $\dfrac{dy}{dx}=\dfrac{1}{dx/dy}$ and expresses the first derivative purely in terms of $y$.

Step 3: Differentiate the first derivative again, treating $y$ as a function of $x$ …

Common Mistakes

Mistake 1: Differentiating $\cos^2 y$ a second time without the chain-rule factor

Why it's wrong: at the second-derivative stage, $y$ is still a function of $x$, so $\dfrac{d}{dx}(\cos^2 y) = 2\cos y(-\sin y)\cdot\dfrac{dy}{dx}$, not just $-2\cos y\sin y$. Leaving out the trailing $\dfrac{dy}{dx}$ gives an expression that isn't actually $\dfrac{d^2y}{dx^2}$. Correct approach: apply the chain rule again exactly as in the first differentiation.

Mistake 2: Never substituting $\dfrac{dy}{dx}=\cos^2 y$ back in

Why it's wrong: the question specifically asks for the answer "in terms of $y$ alone" — stopping at $-2\sin y\cos y\cdot\dfrac{dy}{dx}$ leaves a mixed expression that still contains $\dfrac{dy}{dx}$, not a pure function of $y$. Correct approach: replace $\dfrac{dy}{dx}$ with $\cos^2 y$ (found in step 1) before simplifying to $-2\sin y\cos^3 y$. …