#Derivatives of Inverse Trigonometric Functions

#Table of Contents

1. Introduction
2. Inverse Function Theorem
3. Differentiating Inverse Sine
4. Differentiating Inverse Cosine
5. Differentiating Inverse Tangent
6. Summary of Derivatives
7. Worked Example
8. Practice Questions
9. Glossary
10. Key Takeaways

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#Introduction

In this section, we will explore the differentiation of inverse trigonometric functions. These functions are crucial in calculus and can be differentiated using various methods, including the inverse function theorem, the chain rule, and trigonometric identities.

#Inverse Function Theorem

The inverse function theorem is essential for differentiating inverse trigonometric functions. It can be written as:

$$\frac{dy}{dx} = \frac{1}{\left(\frac{dx}{dy}\right)}$$

or

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

where $g(x) = {f}^{-1}(x)$.

#Differentiating Inverse Sine

To differentiate $\sin^{-1} x$, we use the inverse function theorem:

1. Let $g(x) = \sin^{-1} x$ and $f(x) = \sin x$.
2. Then, ${f}^{\prime}(x) = \cos x$.
3. By the inverse function theorem:

$${g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{\cos(\sin^{-1} x)}$$

We use the identity $\sin^2 x + \cos^2 x \equiv 1$:

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

Therefore:

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

Thus, the derivative of $\sin^{-1} x$ is:

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

#Differentiating Inverse Cosine

To differentiate $\cos^{-1} x$, we follow a similar process:

1. Let $g(x) = \cos^{-1} x$ and $f(x) = \cos x$.
2. Then, ${f}^{\prime}(x) = -\sin x$.
3. By the inverse function theorem:

$${g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{-\sin(\cos^{-1} x)}$$

We use the identity $\sin^2 x + \cos^2 x \equiv 1$:

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

Therefore:

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

Thus, the derivative of $\cos^{-1} x$ is:

$$\frac{d}{dx} (\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}}$$

#Differentiating Inverse Tangent

To differentiate $\tan^{-1} x$, we use the inverse function theorem:

1. Let $g(x) = \tan^{-1} x$ and $f(x) = \tan x$.
2. Then, ${f}^{\prime}(x) = \sec^2 x$.
3. By the inverse function theorem:

$${g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{\sec^2(\tan^{-1} x)}$$

We use the identity $\tan^2 x + 1 \equiv \sec^2 x$:

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

Thus, the derivative of $\tan^{-1} x$ is:

$$\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2}$$

#Summary of Derivatives

The table below summarizes the derivatives of all six inverse trigonometric functions:

#Worked Example

**Find the derivative of $f(x) = \mathrm{arcsin}(2x^3 + e^{2x})$.**

Solution:

Recall that for $\sin^{-1} x$, we can use the chain rule.

Let $y = \sin^{-1}(2x^3 + e^{2x})$ and $u = 2x^3 + e^{2x}$. Then $y = \sin^{-1} u$.

Differentiate both functions:

$$\frac{du}{dx} = 6x^2 + 2e^{2x}$$

$$\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}$$

Apply the chain rule:

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot (6x^2 + 2e^{2x})$$

Substitute $u$ back in:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x^3 + e^{2x})^2}} \cdot (6x^2 + 2e^{2x})$$

Thus,

$$f^{\prime}(x) = \frac{6x^2 + 2e^{2x}}{\sqrt{1 - (2x^3 + e^{2x})^2}}$$

#Practice Questions

Practice Question

1. Differentiate $f(x) = \mathrm{arccos}(x^2)$.

2. Differentiate $f(x) = \mathrm{arctan}(3x + 4)$.

3. Differentiate $f(x) = \mathrm{arcsec}(2x + 1)$.

#Glossary

• Inverse Function Theorem: A method to find the derivative of an inverse function.
• Chain Rule: A rule for differentiating compositions of functions.
• Trigonometric Identities: Equations involving trigonometric functions that are true for all values of the variables.

#Key Takeaways

• The derivatives of inverse trigonometric functions are derived using the inverse function theorem, chain rule, and trigonometric identities.
• Domain restrictions are crucial for ensuring the derivatives are correctly applied.
• Practice differentiating various compositions of inverse trigonometric functions to master the concepts.

#Conclusion

Understanding the derivatives of inverse trigonometric functions is essential in calculus. By mastering the inverse function theorem, chain rule, and trigonometric identities, you can confidently differentiate these functions in various contexts. Practice regularly to reinforce your knowledge and improve your problem-solving skills.