4.7: Inverse Trigonometric Derivatives

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Oct 27, 2024
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- 4.6: Exponential Growth and Decay
- 4.8: Integrals Involving Arctrig Functions

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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

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

$right triangle: Hyp = 1, Opp = x, Adj = root(1-x^2)$

The triangle above demonstrates that

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

Hence

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

Since the

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

We have

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

Exercise

Simplify

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

Derivatives of the Arctrigonometric Functions

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

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

What is

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

We use the formula:

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

Since

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

we have

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

so that

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

Relationships

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

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

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

Recall that

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

hence

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

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

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

Similarly:

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

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

Example 2

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

Solution

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

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

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

We arrive at

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

Contributors and Attributions

Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.

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Example 1

Solution

Exercise

Example 2

Solution

Definition of the Inverse Trig Functions

Inverse of Arctrig Functions

Example 1

Solution

Exercise

Derivatives of the Arctrigonometric Functions

Relationships

Example 2

Solution

Contributors and Attributions

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