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Mathematics LibreTexts

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

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

right triangle:  Hyp = 1, Opp = x, Adj = root(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.