Core Pure · Methods in Calculus · Year 2

Differentiating Inverse Trigonometric Functions

Implicit differentiation unlocks derivatives for arcsin, arccos and arctan.

8 min read 2 worked examples

Differentiating the inverse trig functions

Theorem 3.3 — derivatives of arcsin, arccos, arctan

$$\frac{d}{dx}(\arcsin x)=\frac1{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arccos x)=-\frac1{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arctan x)=\frac1{1+x^2}$$

Notice that $\arcsin x$ and $\arccos x$ have derivatives that are negatives of each other — consistent with $\arcsin x+\arccos x=\dfrac\pi2$ being constant, so their derivatives must sum to zero.

The graph of y=arcsin x: an increasing curve from -pi/2 to pi/2, restricted to the domain -1 to 1

The graph of $y=\arcsin x$, restricted to the domain/range that makes it a genuine function.

The graph of y=arccos x: a decreasing curve from pi to 0, restricted to the domain -1 to 1

The graph of $y=\arccos x$, restricted to the domain/range that makes it a genuine function.

The graph of y=arctan x: an S-shaped curve with horizontal asymptotes at y=pi/2 and y=-pi/2

The graph of $y=\arctan x$, restricted to the domain/range that makes it a genuine function.

ProofDerivative of arcsin x, by implicit differentiation

Let $y=\arcsin x$, so $\sin y=x$, with $-\dfrac\pi2\leq y\leq\dfrac\pi2$.

Rewrite the inverse function as an equation you can differentiate implicitly.

Differentiating both sides with respect to $x$: $\cos y\cdot\dfrac{dy}{dx}=1$, so $\dfrac{dy}{dx}=\dfrac1{\cos y}$.

Implicit differentiation, treating $y$ as a function of $x$.

Since $\cos^2y+\sin^2y=1$, $\cos y=\pm\sqrt{1-\sin^2y}=\pm\sqrt{1-x^2}$.

Use the Pythagorean identity and substitute $\sin y=x$.

For $-\dfrac\pi2\leq y\leq\dfrac\pi2$, $\cos y\geq0$, so we must take the positive root: $\cos y=\sqrt{1-x^2}$.

The restricted range of $\arcsin$ rules out the negative root.

Therefore $\dfrac{dy}{dx}=\dfrac1{\sqrt{1-x^2}}$.

Substitute back into the result from step 2.

∎

Example 1 · Chain rule with an inverse trig function

Given $y=\arctan(3x)$, find $\dfrac{dy}{dx}$.

Let $u=3x$, so $y=\arctan u$. Then $\dfrac{dy}{du}=\dfrac1{1+u^2}$ and $\dfrac{du}{dx}=3$.

By the chain rule: $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac3{1+9x^2}$.

Example 2 · A product involving arcsin

Given $y=x^2\arcsin x$, find $\dfrac{dy}{dx}$.

By the product rule: $\dfrac{dy}{dx}=2x\arcsin x+x^2\times\dfrac1{\sqrt{1-x^2}}=2x\arcsin x+\dfrac{x^2}{\sqrt{1-x^2}}$.

Exam-style questions

Question 1 5 marks

Given $y=\arccos(2x)$, find $\dfrac{dy}{dx}$, and state the values of $x$ for which the derivative is defined.

Let $u=2x$: $\dfrac{dy}{du}=-\dfrac1{\sqrt{1-u^2}}$, $\dfrac{du}{dx}=2$. (M1, A1)

$\dfrac{dy}{dx}=-\dfrac2{\sqrt{1-4x^2}}$. (M1, A1)

Defined for $|2x|\lt1$, i.e. $-\dfrac12\lt x\lt\dfrac12$. (A1)

Question 2 5 marks

Given $y=\arcsin(4x)$, find $\dfrac{dy}{dx}$, and state the values of $x$ for which the derivative is defined.

Let $u=4x$: $\dfrac{dy}{du}=\dfrac1{\sqrt{1-u^2}}$, $\dfrac{du}{dx}=4$. (M1, A1)

$\dfrac{dy}{dx}=\dfrac4{\sqrt{1-16x^2}}$. (M1, A1)

Defined for $|4x|\lt1$, i.e. $-\dfrac14\lt x\lt\dfrac14$. (A1)

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