4. Inverse Trigonometric Differentiation

Video coming soon4 worked examples

Theory

The derivatives of $\sin^{-1}x$ , $\cos^{-1}x$ and $\tan^{-1}x$ are not obvious, but they all follow from one idea about inverse functions. If $y$ is the inverse of some function, we can swap the roles of the variables and use:

\frac{dy}{dx} = \frac{1}{\dfrac{dx}{dy}} \qquad \left(\text{provided } \frac{dx}{dy} \neq 0\right)

Take $y = \tan^{-1}x$ . By definition this means $x = \tan y$ , which is easy to differentiate:

\frac{dx}{dy} = \sec^2 y = 1 + \tan^2 y = 1 + x^2 \quad\implies\quad \frac{dy}{dx} = \frac{1}{1+x^2}

The same argument applied to $\sin^{-1}$ and $\cos^{-1}$ gives the three standard results:

\frac{d}{dx}\bigl(\sin^{-1}x\bigr) = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\bigl(\cos^{-1}x\bigr) = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\bigl(\tan^{-1}x\bigr) = \frac{1}{1+x^2}

When the argument is a function rather than just $x$ , apply the chain rule — replace $x$ by $f(x)$ throughout and multiply by $f'(x)$ :

\frac{d}{dx}\bigl(\tan^{-1}f(x)\bigr) = \frac{f'(x)}{1 + \bigl[f(x)\bigr]^2}

The Golden Rule: to derive one of these, write the inverse statement ( $y = \sin^{-1}x \iff x = \sin y$ ), differentiate that with respect to $y$ , invert it, and finally convert back into terms of $x$ using a Pythagorean identity.

⚠️ Common Examiner Traps

$\sin^{-1}x$ is not $\frac{1}{\sin x}$ : the $-1$ denotes the inverse function. The reciprocal of $\sin x$ is $\operatorname{cosec} x$ .
Losing the minus sign: $\cos^{-1}$ differentiates to the negative of the $\sin^{-1}$ result — that sign is the only difference between them.
Forgetting the chain rule: for $\tan^{-1}(5x^2)$ you must square the whole argument in the denominator — giving $25x^4$ , not $25x^2$ — and multiply by the derivative of the inside.
Choosing the sign of the square root: in the derivation, $\sqrt{1-x^2}$ is taken as positive because the range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , where $\cos y \ge 0$ . State this — it is a marked step.

Worked examples

Example 1

Prove, from the definition of the inverse function, that $\dfrac{d}{dx}\bigl(\sin^{-1}x\bigr) = \dfrac{1}{\sqrt{1-x^2}}$ .

Step 1: Let $y = \sin^{-1}x$ . By the definition of the inverse function this means:

x = \sin y

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

\frac{dx}{dy} = \cos y \quad\implies\quad \frac{dy}{dx} = \frac{1}{\cos y}

Step 3: Convert back to $x$ using $\sin^2 y + \cos^2 y = 1$ :

\cos^2 y = 1 - \sin^2 y = 1 - x^2 \quad\implies\quad \cos y = \pm\sqrt{1-x^2}

Step 4: Choose the sign. The range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , on which $\cos y \ge 0$ , so the positive root is correct:

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}

Example 2

Differentiate (a) $y = \cos^{-1}(2x)$ and (b) $y = \tan^{-1}(5x^2)$ .

Step 1 (a): Apply the standard result for $\cos^{-1}$ — note the minus sign — with the chain rule. The inner function is $2x$ , so multiply by $2$ :

\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (2x)^2}} \times 2

Step 2: Simplify, remembering the whole of $2x$ is squared:

\frac{dy}{dx} = -\frac{2}{\sqrt{1 - 4x^2}}

Step 3 (b): Now the inner function is $5x^2$ , whose derivative is $10x$ :

\frac{dy}{dx} = \frac{1}{1 + (5x^2)^2} \times 10x

Step 4: Square the inner function carefully — $(5x^2)^2 = 25x^4$ , not $25x^2$ :

\frac{dy}{dx} = \frac{10x}{1 + 25x^4}

Example 3

Differentiate $y = \sin^{-1}\!\left(\dfrac{x}{4}\right)$ , simplifying your answer.

Step 1: Use the chain rule with inner function $\dfrac{x}{4}$ , whose derivative is $\dfrac{1}{4}$ :

\frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{x}{4}\right)^2}} \times \frac{1}{4}

Step 2: Simplify the surd by writing the inside as a single fraction:

\sqrt{1 - \frac{x^2}{16}} = \sqrt{\frac{16-x^2}{16}} = \frac{\sqrt{16-x^2}}{4}

Step 3: Substitute back — the factors of $4$ cancel:

\frac{dy}{dx} = \frac{4}{\sqrt{16-x^2}} \times \frac{1}{4} = \frac{1}{\sqrt{16-x^2}}

Example 4

Differentiate $y = x^2\tan^{-1}x$ .

Step 1: This is a product, with $u = x^2$ and $v = \tan^{-1}x$ :

u' = 2x, \qquad v' = \frac{1}{1+x^2}

Step 2: Apply the product rule $u'v + uv'$ :

\frac{dy}{dx} = 2x\tan^{-1}x + x^2 \times \frac{1}{1+x^2}

Step 3: Tidy the second term. The $\tan^{-1}x$ cannot be simplified further, so it stays as it is:

\frac{dy}{dx} = 2x\tan^{-1}x + \frac{x^2}{1+x^2}