Derivatives of inverse trigonometric functions

Because $\sin^{-1}x$, $\cos^{-1}x$, and so on are defined as the inverses of trig functions restricted to a principal-value branch, their derivatives can be found by implicit differentiation of the corresponding direct trig equation. The result is a short, memorisable table.

where: $x$ is the point at which the derivative is evaluated, and the domain restriction next to each formula is where the corresponding inverse function is differentiable — outside it the expression is either undefined or the formula does not apply. Three sub-points are worth holding onto before you touch a single problem.

The co-function pattern. Each "co" function — $\cos^{-1}$, $\cot^{-1}$, $\csc^{-1}$ — has exactly the negative of the derivative of its partner ($\sin^{-1}$, $\tan^{-1}$, $\sec^{-1}$ respectively). If you remember only the first three formulas, you can write down the other three by flipping the sign.

The chain rule wraps every formula. When the argument is a function $g(x)$ rather than plain $x$, for example $y = \sin^{-1}(g(x))$, the derivative picks up a factor of $g'(x)$:

This is the single most common source of lost marks: forgetting the $g'(x)$ factor entirely.

Simplify before you differentiate. A large class of problems present an intimidating argument inside $\sin^{-1}$ or $\tan^{-1}$ that becomes trivial after a trigonometric substitution such as $x = \sin\theta$ or $x = \tan\theta$, using identities like the double-angle formulas. Spotting this pattern before reaching for the chain rule saves several lines of algebra and avoids sign errors.

Setting up the goal. The full goal is to derive $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$ for $-1 < x < 1$ starting only from the definition of $\sin^{-1}x$ as the inverse of sine restricted to the principal branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, using implicit differentiation and the Pythagorean identity — the same method then reproduces all five remaining formulas by an identical pattern applied to $\cos y = x$, $\tan y = x$, and so on.

Step 1. Let $y = \sin^{-1}x$. By definition of the inverse function, this means $\sin y = x$, with $y$ restricted to the principal branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 2. Differentiate both sides of $\sin y = x$ implicitly with respect to $x$, treating $y$ as a function of $x$. The left side needs the chain rule: $\cos y \cdot \dfrac{dy}{dx} = 1$.

Step 3. Solve for $\dfrac{dy}{dx}$: $\dfrac{dy}{dx} = \dfrac{1}{\cos y}$, valid wherever $\cos y \neq 0$, i.e. away from $y = \pm\frac{\pi}{2}$, which corresponds to $x = \pm 1$.

Step 4. Express $\cos y$ in terms of $x$ using $\cos^2 y + \sin^2 y = 1$, so $\cos y = \pm\sqrt{1-\sin^2 y} = \pm\sqrt{1-x^2}$. On the principal branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, cosine is never negative, so we take the positive root: $\cos y = \sqrt{1-x^2}$.

Step 5. Substitute back into Step 3 to get $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}}$ for $-1 < x < 1$, exactly the stated formula. The other five formulas follow from the identical three-step pattern (differentiate implicitly, solve for the derivative in terms of the direct trig function, rewrite that function in terms of $x$ using an identity and the correct sign on the relevant branch) applied to $\cos y = x$, $\tan y = x$, $\cot y = x$, $\sec y = x$, and $\csc y = x$ in turn.

Differentiate $y = \tan^{-1}(x^2)$ with respect to $x$, then extend the method to $y = \sin^{-1}(2x\sqrt{1-x^2})$ using a substitution.

Step 1. For $y = \tan^{-1}(x^2)$, identify the composite structure: outer function $\tan^{-1}(u)$ with inner function $u = x^2$.

Step 2. Differentiate the outer function with respect to $u$ using the standard formula, then substitute $u = x^2$: $\dfrac{dy}{du} = \dfrac{1}{1+u^2} = \dfrac{1}{1+x^4}$.

Step 3. Differentiate the inner function: $\dfrac{du}{dx} = 2x$. By the chain rule, $\dfrac{dy}{dx} = \dfrac{1}{1+x^4}\cdot 2x = \dfrac{2x}{1+x^4}$.

Step 4. Now the second, harder function: $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-\frac{1}{\sqrt2} \le x \le \frac{1}{\sqrt2}$. Substitute $x = \sin\theta$, so $\theta = \sin^{-1}x$ and $\sqrt{1-x^2} = \cos\theta$ for $\theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$. Then $2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta$ by the double-angle identity.

Step 5. Since $\theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, the doubled angle $2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ lies exactly inside the principal branch, so $\sin^{-1}(\sin 2\theta) = 2\theta$ and $y = 2\theta = 2\sin^{-1}x$. Differentiating this simplified form directly gives $\dfrac{dy}{dx} = 2\cdot\dfrac{1}{\sqrt{1-x^2}} = \dfrac{2}{\sqrt{1-x^2}}$.

Final answers: $\dfrac{dy}{dx} = \dfrac{2x}{1+x^4}$ for $y = \tan^{-1}(x^2)$, and $\dfrac{dy}{dx} = \dfrac{2}{\sqrt{1-x^2}}$ for $y = \sin^{-1}(2x\sqrt{1-x^2})$.

Competitive stretch

If instead the problem gave $y = \cos^{-1}(1-2x^2)$ with $x = \sin\theta$, the same substitution turns the argument into $1 - 2\sin^2\theta = \cos 2\theta$, so $y = \cos^{-1}(\cos 2\theta)$. For $\theta \in \left[0, \frac{\pi}{2}\right]$ (i.e. $x \ge 0$), $2\theta \in [0, \pi]$ lies in the principal branch of $\cos^{-1}$, giving $y = 2\theta = 2\sin^{-1}x$, and hence the same derivative $\dfrac{dy}{dx} = \dfrac{2}{\sqrt{1-x^2}}$ — a good check that different-looking arguments can collapse to the same simplified answer.

• Mixing up the sign between a function and its co-function derivative, e.g. writing $\frac{d}{dx}\cos^{-1}x = \frac{1}{\sqrt{1-x^2}}$ instead of the correct negative sign.
• Forgetting the domain restrictions. The formulas for $\sin^{-1}x$ and $\cos^{-1}x$ hold only for $-1<x<1$; for $\sec^{-1}x$ and $\csc^{-1}x$, only for $|x|>1$, and the $|x|$ in the denominator must not be dropped.
• Forgetting to apply the chain rule when the argument is not simply $x$, e.g. differentiating $\sin^{-1}(3x)$ as $\frac{1}{\sqrt{1-x^2}}$ instead of $\frac{3}{\sqrt{1-9x^2}}$.
• Choosing the wrong substitution range, so the simplified angle falls outside the principal-value branch, making a cancellation like $\sin^{-1}(\sin 2\theta) = 2\theta$ invalid — always check the branch before dropping the outer and inner inverse functions.

Ask the tutor: Want to practice picking the right substitution for a gnarly inverse-trig argument, with guided hints on tracking the principal-value branch?