Derivatives of inverse trigonometric functions

This lesson differentiates three inverse trigonometric functions. In each case we first make a substitution that turns the inside expression into a double-angle identity, simplify the function down to a constant times an inverse trig function, and then differentiate.

Problem 1: $y = \cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right)$, for $-1 < x < 1$

Substitute $x = \tan\theta$, so that $\theta = \tan^{-1}x$. Then

$y = \cos^{-1}\!\left(\frac{2\tan\theta}{1+\tan^2\theta}\right).$

Using the identity $\dfrac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$,

$y = \cos^{-1}(\sin 2\theta).$

To change the sine into a cosine, use $\sin x = \cos\!\left(\tfrac{\pi}{2} - x\right)$:

$y = \cos^{-1}\!\left(\cos\!\left(\tfrac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta.$

Since $\theta = \tan^{-1}x$,

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

Differentiating with respect to $x$:

$\frac{dy}{dx} = 0 - 2\cdot\frac{1}{1+x^2} = -\frac{2}{1+x^2}.$

Problem 2: $y = \sin^{-1}\!\left(2x\sqrt{1-x^2}\right)$, for $-\tfrac{1}{\sqrt{2}} < x < \tfrac{1}{\sqrt{2}}$

Substitute $x = \sin\theta$, so that $\theta = \sin^{-1}x$. Then

$y = \sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^2\theta}\right) = \sin^{-1}\!\left(2\sin\theta\sqrt{\cos^2\theta}\right) = \sin^{-1}(2\sin\theta\cos\theta).$

Since $2\sin\theta\cos\theta = \sin 2\theta$,

$y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x.$

Differentiating with respect to $x$:

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

Problem 3: $y = \cos^{-1}\!\left(2x^2 - 1\right)$, for $0 < x < \tfrac{1}{\sqrt{2}}$

Substitute $x = \cos\theta$, so that $\theta = \cos^{-1}x$. Using the identity $2\cos^2\theta - 1 = \cos 2\theta$,

$y = \cos^{-1}(2\cos^2\theta - 1) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\cos^{-1}x.$

Differentiating with respect to $x$:

$\frac{dy}{dx} = 2\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{2}{\sqrt{1-x^2}}.$

Key takeaways

• A substitution like $x = \tan\theta$, $x = \sin\theta$, or $x = \cos\theta$ exposes a double-angle identity that collapses the expression.
• After simplifying, each function becomes a constant multiple of $\tan^{-1}x$, $\sin^{-1}x$, or $\cos^{-1}x$, which is easy to differentiate.
• The three derivatives are $-\dfrac{2}{1+x^2}$, $\dfrac{2}{\sqrt{1-x^2}}$, and $-\dfrac{2}{\sqrt{1-x^2}}$ respectively.