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# Derivatives of Inverse Trigonometric Functions

### Learning Objectives

• Recognize the derivatives of the standard inverse trigonometric functions.

# Derivatives of Inverse Trigonometric Functions

We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.

### Derivative of the Inverse Sine Function

Use the inverse function theorem to find the derivative of $g(x)=\sin^{-1} x$.

Since for $x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}], \, f(x)= \sin x$ is the inverse of $g(x)= \sin^{-1} x$, begin by finding $f^{\prime}(x)$. Since

$f^{\prime}(x)= \cos x$ and $f^{\prime}(g(x))= \cos (\sin^{-1} x)=\sqrt{1-x^2}$,

we see that

$g^{\prime}(x)=\frac{d}{dx}(\sin^{-1} x)=\frac{1}{f^{\prime}(g(x))}=\frac{1}{\sqrt{1-x^2}}$.

#### Analysis

To see that $\cos (\sin^{-1} x)=\sqrt{1-x^2}$, consider the following argument. Set $\sin^{-1} x=\theta$. In this case, $\sin \theta =x$ where $-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$. We begin by considering the case where $0<\theta <\frac{\pi}{2}$. Since $\theta$ is an acute angle, we may construct a right triangle having acute angle $\theta$, a hypotenuse of length 1, and the side opposite angle $\theta$ having length $x$. From the Pythagorean theorem, the side adjacent to angle $\theta$ has length $\sqrt{1-x^2}$. This triangle is shown in (Figure). Using the triangle, we see that $\cos (\sin^{-1} x)= \cos \theta =\sqrt{1-x^2}$.

Figure 2. Using a right triangle having acute angle $\theta$, a hypotenuse of length 1, and the side opposite angle $\theta$ having length $x$, we can see that $\cos (\sin^{-1} x)= \cos \theta =\sqrt{1-x^2}$.
In the case where $-\frac{\pi}{2}<\theta <0$, we make the observation that $0<-\theta<\frac{\pi}{2}$ and hence $\cos (\sin^{-1} x)= \cos \theta = \cos (−\theta )=\sqrt{1-x^2}$.

Now if $\theta =\frac{\pi}{2}$ or $\theta =-\frac{\pi}{2}, \, x=1$ or $x=-1$, and since in either case $\cos \theta =0$ and $\sqrt{1-x^2}=0$, we have

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

Consequently, in all cases, $\cos (\sin^{-1} x)=\sqrt{1-x^2}$.

### Applying the Chain Rule to the Inverse Sine Function

Apply the chain rule to the formula derived in (Figure) to find the derivative of $h(x)=\sin^{-1} (g(x))$ and use this result to find the derivative of $h(x)=\sin^{-1}(2x^3)$.

Applying the chain rule to $h(x)=\sin^{-1} (g(x))$, we have

$h^{\prime}(x)=\frac{1}{\sqrt{1-(g(x))^2}}g^{\prime}(x)$.

Now let $g(x)=2x^3$, so $g^{\prime}(x)=6x$. Substituting into the previous result, we obtain

$\begin{array}{ll} h^{\prime}(x) & =\frac{1}{\sqrt{1-4x^6}} \cdot 6x \\ & =\frac{6x}{\sqrt{1-4x^6}}. \end{array}$

Use the inverse function theorem to find the derivative of $g(x)=\tan^{-1} x$.

$g^{\prime}(x)=\frac{1}{1+x^2}$

#### Hint

The inverse of $g(x)$ is $f(x)= \tan x$. Use (Figure) as a guide.

The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.

### Derivatives of Inverse Trigonometric Functions

$\begin{array}{lllll}\frac{d}{dx}(\sin^{-1} x)=\large \frac{1}{\sqrt{1-x^2}} & & & & \frac{d}{dx}(\cos^{-1} x)=\large \frac{-1}{\sqrt{1-x^2}} \\ \frac{d}{dx}(\tan^{-1} x)=\large \frac{1}{1+x^2} & & & & \frac{d}{dx}(\cot^{-1} x)=\large \frac{-1}{1+x^2} \\ \frac{d}{dx}(\sec^{-1} x)=\large \frac{1}{|x|\sqrt{x^2-1}} & & & & \frac{d}{dx}(\csc^{-1} x)=\large \frac{-1}{|x|\sqrt{x^2-1}} \end{array}$

### Applying Differentiation Formulas to an Inverse Tangent Function

Find the derivative of $f(x)=\tan^{-1} (x^2)$.

Let $g(x)=x^2$, so $g^{\prime}(x)=2x$. Substituting into (Figure), we obtain

$f^{\prime}(x)=\frac{1}{1+(x^2)^2} \cdot (2x)$.

Simplifying, we have

$f^{\prime}(x)=\frac{2x}{1+x^4}$.

### Applying Differentiation Formulas to an Inverse Sine Function

Find the derivative of $h(x)=x^2 \sin^{-1} x$.

By applying the product rule, we have

$h^{\prime}(x)=2x \sin^{-1} x+\frac{1}{\sqrt{1-x^2}} \cdot x^2$.

Find the derivative of $h(x)= \cos^{-1} (3x-1)$.

$h^{\prime}(x)=\frac{-3}{\sqrt{6x-9x^2}}$

#### Hint

Use (Figure). with $g(x)=3x-1$

### Applying the Inverse Tangent Function

The position of a particle at time $t$ is given by $s(t)= \tan^{-1}(\frac{1}{t})$ for $t\ge \frac{1}{2}$. Find the velocity of the particle at time $t=1$.

Begin by differentiating $s(t)$ in order to find $v(t)$. Thus,

$v(t)=s^{\prime}(t)=\frac{1}{1+(\frac{1}{t})^2} \cdot \frac{-1}{t^2}$.

Simplifying, we have

$v(t)=-\frac{1}{t^2+1}$.

Thus, $v(1)=-\frac{1}{2}$.

Find the equation of the line tangent to the graph of $f(x)= \sin^{-1} x$ at $x=0$.

$y=x$

#### Hint

$f^{\prime}(0)$ gives the slope of the tangent line.

### Key Concepts

• We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.

# Key Equations

• Derivative of inverse sine function $\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}$
• Derivative of inverse cosine function $\frac{d}{dx}(\cos^{-1} x)=\frac{-1}{\sqrt{1-x^2}}$
• Derivative of inverse tangent function $\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}$
• Derivative of inverse cotangent function $\frac{d}{dx}(\cot^{-1} x)=\frac{-1}{1+x^2}$
• Derivative of inverse secant function $\frac{d}{dx}(\sec^{-1} x)=\frac{1}{|x|\sqrt{x^2-1}}$
• Derivative of inverse cosecant function $\frac{d}{dx}(\csc^{-1} x)=\frac{-1}{|x|\sqrt{x^2-1}}$

For the following exercises, find $\frac{dy}{dx}$ for the given function.

1. $y= \sin^{-1}(x^2)$

$\large \frac{2x}{\sqrt{1-x^4}}$

2. $y= \cos^{-1}(\sqrt{x})$

3. $y= \sec^{-1}(\frac{1}{x})$

$\large \frac{-1}{\sqrt{1-x^2}}$

4. $y=\sqrt{\csc^{-1} x}$

5. $y=(1 + \tan^{-1} x)^3$

$\large \frac{3(1 + \tan^{-1} x)^2}{1+x^2}$

6. $y= \cos^{-1}(2x) \cdot \sin^{-1}(2x)$

7. $y=\frac{1}{\tan^{-1}(x)}$

$\large \frac{-1}{(1+x^2)(\tan^{-1} x)^2}$

8. $y= \sec^{-1}(−x)$

9. $y= \cot^{-1} \sqrt{4-x^2}$

$\large \frac{x}{(5-x^2)\sqrt{4-x^2}}$

10. $y=x \cdot \csc^{-1} x$

11. [T] The position of a moving hockey puck after $t$ seconds is $s(t)= \tan^{-1} t$ where $s$ is in meters.

1. Find the velocity of the hockey puck at any time $t$.
2. Find the acceleration of the puck at any time $t$.
3. Evaluate a. and b. for $t=2,4$, and 6 seconds.
4. What conclusion can be drawn from the results in c.?

a. $v(t)=\frac{1}{1+t^2}$ b. $a(t)=\frac{-2t}{(1+t^2)^2}$ c. $v(2)=0.2, \, v(4)=0.06, \, v(6)=0.03; \, a(2)=-0.16, \, a(4)=-0.028, \, a(6)=-0.0088$ d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.