zuai-logo

Differentiation of Composite & Inverse Functions

Emily Davis

Emily Davis

6 min read

Listen to this study note

Study Guide Overview

This study guide covers the derivatives of inverse trigonometric functions. It explains the inverse function theorem and how it's used to find the derivatives of arcsin(x), arccos(x), and arctan(x). It summarizes the derivatives of all six inverse trig functions and provides a worked example and practice problems. Key terms like the chain rule and trigonometric identities are also explained.

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

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:

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

or

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

where g(x)=f1(x)g(x) = {f}^{-1}(x).

Key Concept

Inverse trigonometric functions are sometimes referred to using "arc" notation, e.g., arcsin,x\mathrm{arcsin} , x is the same as sin1x\sin^{-1} x.

Differentiating Inverse Sine

To differentiate sin1x\sin^{-1} x, we use the inverse function theorem:

  1. Let g(x)=sin1xg(x) = \sin^{-1} x and f(x)=sinxf(x) = \sin x.
  2. Then, f(x)=cosx{f}^{\prime}(x) = \cos x.
  3. By the inverse function theorem:

g(x)=1f(g(x))=1cos(sin1x){g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{\cos(\sin^{-1} x)}

We use the identity sin2x+cos2x1\sin^2 x + \cos^2 x \equiv 1:

cosx=1sin2x\cos x = \sqrt{1 - \sin^2 x}

Therefore:

g(x)=11x2{g}^{\prime}(x) = \frac{1}{\sqrt{1 - x^2}}

Thus, the derivative of sin1x\sin^{-1} x is:

ddx(sin1x)=11x2\frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}

Common Mistake

Ensure to use the domain restrictions for sinx\sin x and sin1x\sin^{-1} x. The derivative is defined only for 1<x<1-1 < x < 1.

Differentiating Inverse Cosine

To differentiate cos1x\cos^{-1} x, we follow a similar process:

  1. Let g(x)=cos1xg(x) = \cos^{-1} x and f(x)=cosxf(x) = \cos x.
  2. Then, f(x)=sinx{f}^{\prime}(x) = -\sin x.
  3. By the inverse function theorem:

g(x)=1f(g(x))=1sin(cos1x){g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{-\sin(\cos^{-1} x)}

We use the identity sin2x+cos2x1\sin^2 x + \cos^2 x \equiv 1:

sinx=1cos2x\sin x = \sqrt{1 - \cos^2 x}

Therefore:

g(x)=11x2{g}^{\prime}(x) = \frac{1}{-\sqrt{1 - x^2}}

Thus, the derivative of cos1x\cos^{-1} x is:

ddx(cos1x)=11x2\frac{d}{dx} (\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}}

Common Mistake

Ensure to use the domain restrictions for cosx\cos x and cos1x\cos^{-1} x. The derivative is defined only for 1<x<1-1 < x < 1.

Differentiating Inverse Tangent

To differentiate tan1x\tan^{-1} x, we use the inverse function theorem:

  1. Let g(x)=tan1xg(x) = \tan^{-1} x and f(x)=tanxf(x) = \tan x.
  2. Then, f(x)=sec2x{f}^{\prime}(x) = \sec^2 x.
  3. By the inverse function theorem:

g(x)=1f(g(x))=1sec2(tan1x){g}^{\prime}(x) = \frac{1}{{f}^{\prime}(g(x))} = \frac{1}{\sec^2(\tan^{-1} x)}

We use the identity tan2x+1sec2x\tan^2 x + 1 \equiv \sec^2 x:

g(x)=11+x2{g}^{\prime}(x) = \frac{1}{1 + x^2}

Thus, the derivative of tan1x\tan^{-1} x is:

ddx(tan1x)=11+x2\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2}

Common Mistake

Ensure to use the domain restrictions for tanx\tan x and tan1x\tan^{-1} x. The derivative is defined for all real numbers.

Summary of Derivatives

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

FunctionDerivative
sin1x\sin^{-1} x11x2\frac{1}{\sqrt{1 - x^2}}
cos1x\cos^{-1} x11x2-\frac{1}{\sqrt{1 - x^2}}
tan1x\tan^{-1} x11+x2\frac{1}{1 + x^2}
csc1x\csc^{-1} x-\frac{1}{
sec1x\sec^{-1} x\frac{1}{
cot1x\cot^{-1} x11+x2-\frac{1}{1 + x^2}

Worked Example

**Find the derivative of f(x)=arcsin(2x3+e2x)f(x) = \mathrm{arcsin}(2x^3 + e^{2x}).**

Solution:

Recall that for sin1x\sin^{-1} x, we can use the chain rule.

Let y=sin1(2x3+e2x)y = \sin^{-1}(2x^3 + e^{2x}) and u=2x3+e2xu = 2x^3 + e^{2x}. Then y=sin1uy = \sin^{-1} u.

Differentiate both functions:

dudx=6x2+2e2x\frac{du}{dx} = 6x^2 + 2e^{2x}

dydu=11u2\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}

Apply the chain rule:

dydx=dydu×dudx=11u2(6x2+2e2x)\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot (6x^2 + 2e^{2x})

Substitute uu back in:

dydx=11(2x3+e2x)2(6x2+2e2x)\frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x^3 + e^{2x})^2}} \cdot (6x^2 + 2e^{2x})

Thus,

f(x)=6x2+2e2x1(2x3+e2x)2f^{\prime}(x) = \frac{6x^2 + 2e^{2x}}{\sqrt{1 - (2x^3 + e^{2x})^2}}

Practice Questions

Practice Question

1. Differentiate f(x)=arccos(x2)f(x) = \mathrm{arccos}(x^2).

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

3. Differentiate f(x)=arcsec(2x+1)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.

Question 1 of 11

What is the derivative of sin1(x)\sin^{-1}(x)? 🤔

11x2\frac{1}{\sqrt{1 - x^2}}

11x2-\frac{1}{\sqrt{1 - x^2}}

11+x2\frac{1}{1 + x^2}

11+x2-\frac{1}{1 + x^2}