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Differentiation of Composite & Inverse Functions

Emily Davis

Emily Davis

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Next Topic - Derivatives of Inverse Trigonometric Functions

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Study Guide Overview

This study guide covers the reciprocal of a derivative and derivatives of inverse functions. It explains the relationship between f'(x) and (f⁻¹)'(x), including the derivation and application of the Inverse Function Theorem. The guide provides worked examples, practice questions, and a glossary of key terms like derivative, inverse function, and chain rule. Key concepts include finding the derivative of an inverse function and relating rates of change.

#Study Notes: The Reciprocal of a Derivative and Derivatives of Inverse Functions

#Table of Contents

  1. The Reciprocal of a Derivative
    • Concept Overview
    • Useful Applications
  2. Derivatives of Inverse Functions
    • Relationship between Function and Inverse Derivatives
    • Deriving the Inverse Function Theorem
  3. Worked Examples
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

#The Reciprocal of a Derivative

#Concept Overview

Derivatives are not fractions, but they behave similarly when finding reciprocals. The reciprocal of a derivative is crucial in various calculations, especially involving inverse functions.

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

This holds true as long as dydx≠0\frac{dy}{dx} \neq 0dxdy​=0.

Likewise, the reciprocal of dxdy\frac{dx}{dy}dydx​ is:

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

This is valid if dxdy≠0\frac{dx}{dy} \neq 0dydx​=0.

#Useful Applications

Key Concept
  • Finding the derivative of the inverse of a function
  • Relating rates of change using the chain rule

#Derivatives of Inverse Functions

#Relationship between Function and Inverse Derivatives

For a function fff that is differentiable and has an inverse f−1f^{-1}f−1, the derivative of the inverse at a point aaa is given by:

(f−1)′(a)=1f′(f−1(a))\left( f^{-1} \right)'(a) = \frac{1}{f'\left( f^{-1}(a) \right)}(f−1)′(a)=f′(f−1(a))1​

This is valid if f′(f−1(a))≠0f'\left( f^{-1}(a) \right) \neq 0f′(f−1(a))=0.

You may also see this written as:

g′(a)=1f′(g(a))g'(a) = \frac{1}{f'\left( g(a) \right)}g′(a)=f′(g(a))1​

Where g(a)=f−1(a)g(a) = f^{-1}(a)g(a)=f−1(a).

This property is known as the Inverse Function Theorem.

#Deriving the Inverse Function Theorem

To derive the Inverse Function Theorem, we use the definition of an inverse and the chain rule:

  1. Let g(x)=f−1(x)g(x) = f^{-1}(x)g(x)=f−1(x).
  2. Since fff and ggg are inverses, we have f(g(x))=xf(g(x)) = xf(g(x))=x.
  3. Differentiate both sides with respect to xxx:

ddx(f(g(x)))=ddx(x)\frac{d}{dx} \left( f(g(x)) \right) = \frac{d}{dx} (x)dxd​(f(g(x)))=dxd​(x)

  1. Apply the chain rule to the left-hand side:

f′(g(x))⋅g′(x)=1f'\left( g(x) \right) \cdot g'(x) = 1f′(g(x))⋅g′(x)=1

  1. Rearrange to solve for g′(x)g'(x)g′(x):

g′(x)=1f′(g(x))g'(x) = \frac{1}{f'\left( g(x) \right)}g′(x)=f′(g(x))1​

  1. Recall that g(x)=f−1(x)g(x) = f^{-1}(x)g(x)=f−1(x):

(f−1)′(x)=1f′(f−1(x))\left( f^{-1} \right)'(x) = \frac{1}{f'\left( f^{-1}(x) \right)}(f−1)′(x)=f′(f−1(x))1​


#Worked Examples

#Example 1

Given that f(0)=256f(0) = 256f(0)=256, find the value of g′(256)g'(256)g′(256).

Write the inverse function theorem using f(x)f(x)f(x) and f−1(x)=g(x)f^{-1}(x) = g(x)f−1(x)=g(x):

g′(256)=1f′(g(256))g'(256) = \frac{1}{f'\left( g(256) \right)}g′(256)=f′(g(256))1​

Since f(0)=256f(0) = 256f(0)=256, then g(256)=0g(256) = 0g(256)=0. Now find f′(0)f'(0)f′(0):

f′(x)=12(3x+4)3f'(x) = 12(3x + 4)^3f′(x)=12(3x+4)3

Evaluate at x=0x = 0x=0:

f′(0)=12(3(0)+4)3=12(4)3=768f'(0) = 12(3(0) + 4)^3 = 12(4)^3 = 768f′(0)=12(3(0)+4)3=12(4)3=768

So,

g′(256)=1768g'(256) = \frac{1}{768}g′(256)=7681​

#Example 2

Show that the inverse of f(x)=x3+2x−10f(x) = x^3 + 2x - 10f(x)=x3+2x−10 exists and find the derivative of the inverse at x=125x = 125x=125.

First, check if the inverse exists:

f′(x)=3x2+2f'(x) = 3x^2 + 2f′(x)=3x2+2

Since f′(x)f'(x)f′(x) is always positive, f(x)f(x)f(x) is one-to-one and thus f−1(x)f^{-1}(x)f−1(x) exists.

Use the inverse function theorem:

(f−1)′(125)=1f′(f−1(125))\left( f^{-1} \right)'(125) = \frac{1}{f'\left( f^{-1}(125) \right)}(f−1)′(125)=f′(f−1(125))1​

Find f−1(125)f^{-1}(125)f−1(125). Solve:

125=y3+2y−10125 = y^3 + 2y - 10125=y3+2y−10

By solving, y=5y = 5y=5:

f−1(125)=5f^{-1}(125) = 5f−1(125)=5

Thus,

(f−1)′(125)=1f′(5)\left( f^{-1} \right)'(125) = \frac{1}{f'(5)}(f−1)′(125)=f′(5)1​

Evaluate f′(5)f'(5)f′(5):

f′(5)=3(5)2+2=77f'(5) = 3(5)^2 + 2 = 77f′(5)=3(5)2+2=77

So,

(f−1)′(125)=177\left( f^{-1} \right)'(125) = \frac{1}{77}(f−1)′(125)=771​


#Practice Questions

Practice Question
  1. Given f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1, find (f−1)′(2)\left( f^{-1} \right)'(2)(f−1)′(2).
Practice Question
  1. If h(x)=ln⁡(x)h(x) = \ln(x)h(x)=ln(x), find (h−1)′(e)\left( h^{-1} \right)'(e)(h−1)′(e).
Practice Question
  1. Show that the inverse of f(x)=ex+xf(x) = e^x + xf(x)=ex+x exists and find (f−1)′(2)\left( f^{-1} \right)'(2)(f−1)′(2).

#Glossary

  • Derivative: A measure of how a function changes as its input changes.
  • Inverse Function: A function that "reverses" another function.
  • Chain Rule: A formula for computing the derivative of the composition of two or more functions.
  • Inverse Function Theorem: A theorem that provides a formula for the derivative of the inverse of a function.

#Summary and Key Takeaways

#Summary

  • The reciprocal of a derivative dydx\frac{dy}{dx}dxdy​ is dxdy\frac{dx}{dy}dydx​.
  • The Inverse Function Theorem provides a method to find the derivative of the inverse function.
  • The theorem is derived using the chain rule and the definition of an inverse function.

#Key Takeaways

  • Understand the reciprocal relationship between dydx\frac{dy}{dx}dxdy​ and dxdy\frac{dx}{dy}dydx​.
  • Apply the Inverse Function Theorem to find the derivative of inverse functions.
  • Use the chain rule effectively in derivations and calculations.
Exam Tip

Remember, always verify if the function is one-to-one before applying the Inverse Function Theorem.

Common Mistake

A common mistake is to forget that the reciprocal relationship only holds when the derivative is non-zero.

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Question 1 of 10

Let's warm up! 🚀 If dydx=5\frac{dy}{dx} = 5dxdy​=5 at a certain point, what is the value of dxdy\frac{dx}{dy}dydx​ at that same point?

5

-5

15\frac{1}{5}51​

0