Course Content
AP Calculus BC

Derivatives of Inverse Trigonometric Functions


Introduction

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions.
They are used to find the angle when the ratio of sides in a triangle is known.

If: $$y=\sin(x)$$

then inverse is : $$x = \sin(y)$$

These functions are important in calculus, geometry, and physics because they allow us to reverse trigonometric relationships and differentiate or integrate them.


Using Implicit Differentiation

Let’s take $$y = \sin^\prime(x)$$. then: $$\sin(y) = x$$

Differentiate both sides with respect to xx:

$$\sin^2(y) + \cos^2(y) = 1$$ here we can have $$\cos(y) = \sqrt{1-sin^{2}(y)}$$

and $$\sin(y) = x$$


List of Derivatives

Function Derivative Domain
$$\frac{d}{dx}\sin^{-1}(u)$$ $$\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}$$ $$-1<x<1$$
$$\frac{d}{dx}\cos^{-1}(u)$$ $$-1<x<1$$
$$\frac{d}{dx}\tan^{-1}(u)$$ all real x
$$\frac{d}{dx}\cot^{-1}(u)$$ all real x
$$\frac{d}{dx}\sec^{-1}(u)$$ $$\frac{1}{|u|\sqrt{u^{2}-1}}$$ $$x\not=\pm1,0$$
$$\frac{d}{dx}\csc^{-1}(u)$$ $$\frac{-1}{|u|\sqrt{u^{2}-1}}$$ $$x\not=\pm1,0$$

Example 1


Example 2

$$f^\prime(x) = \frac{-1}{\sqrt{1-(2x)^2}}\cdot(2x)^\prime=\frac{2}{\sqrt{1-4x^2}}$$


Example 3

$$y^\prime=\frac{1}{1+(x^2)^2}\cdot(x^2)^\prime=\frac{2x}{1+x^4}$$