
# 4.7: Inverse Trigonometric Derivatives

### Definition of the Inverse Trig Functions

Recall that we write $$\sin^{-1} x$$ or $$\text{arcsin}\, x$$ to mean the inverse $$\sin$$ of $$x$$ restricted to have values between $$-\pi/2$$ and $$\pi/2$$ (Note that $$\sin x$$ does not pass the horizontal line test, hence we need to restrict the domain.) We define the other five inverse trigonometric functions similarly.

### Inverse of Arctrig Functions

Example 1

Find $$\tan(\sin^{-1} x)$$

Solution

$\tan(\sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.$

The triangle above demonstrates that

$\sin t = \dfrac{x}{1} = \dfrac{opp}{hyp}.$

Hence

$\tan(\tan^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.$

Since the

$\text{tangent} = \dfrac{opp}{adj}.$

We have

$\tan ( \sin^{-1} x) = \dfrac{x}{\sqrt{1-x^2}}.$

Exercise

Simplify

$\cos(\tan^{-1} (2x)).$

### Derivatives of the Arctrigonometric Functions

Recall that if $$f$$ and $$g$$ are inverses, then

$g'(x) \dfrac{1}{f'(g(x))}.$

What is

$\dfrac{d}{dx} \tan^{-1} x\text{?}$

We use the formula:

$\frac{d}{dx} \tan^{-1} x= \dfrac{1}{\sec^2 (\tan^{-1} x)} = \cos^2 (\tan^{-1} x).$

Since

$\tan q = \dfrac{opp}{adj} = \dfrac{x}{1}$

we have

$hyp = \sqrt{1+x^2}$

so that

$\cos^2 (\tan^{-1} x) = \left( \dfrac{1}{\sqrt{1+x^2}}\right)^2 = \dfrac{1}{1+x^2}.$

### Relationships

$\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}$

$\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$

$\dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{|x|\sqrt{x^2-1|}}$

Recall that

$\cos x = \sin \left(\dfrac{\pi}{2} - x \right)$

hence

$\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$

so

$\dfrac{d}{dx} \cos^{-1} x = \dfrac{d}{dx} \left[ \dfrac{\pi}{2} - \sin^{-1} x \right]$

$= \dfrac{-d}{dx} \sin^{-1} x = \dfrac{-1}{\sqrt{1-x^2}}.$

Similarly:

$\dfrac{d}{dx} \cot^{-1} x = \dfrac{-1}{\sqrt{1 + x^2}}$

$\dfrac{d}{dx} \text{csc}\, x\ = \dfrac{-1}{\sqrt{1-x^2}}.$

Example 2

Find the derivative of $$\cos(\sin^{-1} x)$$.

Solution

Let $$y = \cos u$$ , $$u = \sin^{-1} x$$, and $$y' = -\sin u$$

$y'u= -\sin (\sin^{-1} x) = x$

$u' = \dfrac{1}{\sqrt{1-x^2}}.$

We arrive at

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

### Contributors

• Integrated by Justin Marshall.