Slide 1

  • Topic: Derivatives of Inverse Functions
  • Introduction to the concept of inverse functions
  • Importance of studying derivatives of inverse functions

Slide 2

  • Recap: Derivatives of basic functions
    • Derivative of a constant function
    • Derivative of a power function
    • Derivative of an exponential function
    • Derivative of a logarithmic function
    • Derivative of a trigonometric function

Slide 3

  • Definition of inverse functions
    • One-to-one functions and their inverses
    • Notation: f(x) and f^(-1)(x)

Slide 4

  • Derivatives of inverse functions
    • Theorem: if f(x) has an inverse function f^(-1)(x), then:
      • f^(-1)’(x) = 1 / f’(f^(-1)(x))

Slide 5

  • Proof of the derivative formula for inverse functions
  • Example: Find the derivative of f^(-1)(x) if f(x) = 2x + 3

Slide 6

  • Example: Find the derivative of f^(-1)(x) if f(x) = sqrt(x)

Slide 7

  • Example: Find the derivative of f^(-1)(x) if f(x) = e^x

Slide 8

  • Example: Find the derivative of f^(-1)(x) if f(x) = ln(x)

Slide 9

  • Example: Find the derivative of f^(-1)(x) if f(x) = sin(x)

Slide 10

  • Example: Find the derivative of f^(-1)(x) if f(x) = cos(x)

Slide 11

  • Derivatives - Examples of Derivatives of Inverse Functions
  • Example 1:
    • Let f(x) = x^3 + 1. Find the derivative of f^(-1)(x) at x = 2.
    • Solution:
      • First, find the inverse function: y = x^3 + 1, then solve for x.
      • y = f(x) = x^3 + 1, x = (y - 1)^(1/3).
      • Now, differentiate both sides with respect to y and solve for dy/dx.
      • dy/dx = 1 / f’(f^(-1)(x))
      • Evaluate f’(f^(-1)(x)) by substituting x = 2: f’(2) = 3(2^2) = 12.
      • Therefore, dy/dx = 1 / 12.
  • Example 2:
    • Let f(x) = sin(x). Find the derivative of f^(-1)(x) at x = 0.
    • Solution:
      • First, find the inverse function: y = sin(x), then solve for x.
      • y = f(x) = sin(x), x = arcsin(y).
      • Now, differentiate both sides with respect to y and solve for dy/dx.
      • dy/dx = 1 / f’(f^(-1)(x))
      • Evaluate f’(f^(-1)(x)) by substituting x = 0: f’(0) = cos(0) = 1.
      • Therefore, dy/dx = 1 / 1.

Slide 12

  • Example 3:
    • Let f(x) = ln(x). Find the derivative of f^(-1)(x) at x = 2.
    • Solution:
      • First, find the inverse function: y = ln(x), then solve for x.
      • y = f(x) = ln(x), x = e^y.
      • Now, differentiate both sides with respect to y and solve for dy/dx.
      • dy/dx = 1 / f’(f^(-1)(x))
      • Evaluate f’(f^(-1)(x)) by substituting x = 2: f’(2) = 1/2.
      • Therefore, dy/dx = 1 / (1/2) = 2.
  • Example 4:
    • Let f(x) = 2x^2 + 3x. Find the derivative of f^(-1)(x) at x = 1.
    • Solution:
      • First, find the inverse function: y = 2x^2 + 3x, then solve for x.
      • y = f(x) = 2x^2 + 3x, 0 = 2x^2 + 3x - y.
      • Use the quadratic formula to solve for x:
        • x = (-3 ± sqrt(3^2 - 4(2)(-y))) / (2(2))
        • x = (-3 ± sqrt(9 + 8y)) / 4
      • Now, differentiate both sides with respect to y and solve for dy/dx.
      • dy/dx = 1 / f’(f^(-1)(x))
      • Evaluate f’(f^(-1)(x)) by substituting x = 1: f’(1) = 7.
      • Therefore, dy/dx = 1 /7.

Slide 13

  • The Chain Rule for Inverse Functions
  • The chain rule states that if f and g are functions such that f(g(x)) = x for all x in the domain of g, then:
    • (f(g(x)))’ = f’(g(x)) * g’(x)
  • Using the chain rule for inverse functions, if f and g are inverse functions:
    • (f^(-1)(f(x)))’ = f^(-1)’(f(x)) * f’(x)
    • (f(f^(-1)(x)))’ = f’(f^(-1)(x)) * (f^(-1))’(x)

Slide 14

  • Proof of the Chain Rule for Inverse Functions
  • Using the definition of inverse functions and the chain rule, we can prove the chain rule for inverse functions.
  • Proof of (f(f^(-1)(x)))’ = f’(f^(-1)(x)) * (f^(-1))’(x)
    • Let y = f^(-1)(x)
    • Then, x = f(y)
    • Differentiating both sides with respect to x:
      • 1 = f’(y) * (f^(-1))’(x)
    • Solving for (f^(-1))’(x):
      • (f^(-1))’(x) = 1 / f’(y)
      • Substituting y = f^(-1)(x):
      • (f^(-1))’(x) = 1 / f’(f^(-1)(x))

Slide 15

  • Example:
    • Let f(x) = 3x + 1 and g(x) = (x - 1)/3 be inverse functions.
    • Find the derivative of (g(f(x)))’ at x = 2.
    • Solution:
      • First, find the composition of the two functions:
        • g(f(x)) = (f(x) - 1)/3 = ((3x + 1) - 1)/3 = x
      • Now, differentiate both sides with respect to x:
        • (g(f(x)))’ = 1

Slide 16

  • Example:
    • Let f(x) = sqrt(x) and g(x) = x^2 be inverse functions.
    • Find the derivative of (f(g(x)))’ at x = 3.
    • Solution:
      • First, find the composition of the two functions:
        • f(g(x)) = sqrt(x^2) = |x|
      • Now, differentiate both sides with respect to x:
        • (f(g(x)))’ = |(x)’| = 1 for x > 0 = |(-x)’| = 1 for x < 0

Slide 17

  • Example:
    • Let f(x) = e^x and g(x) = ln(x) be inverse functions.
    • Find the derivative of (g(f(x)))’ at x = 1.
    • Solution:
      • First, find the composition of the two functions:
        • g(f(x)) = ln(e^x) = x
      • Now, differentiate both sides with respect to x:
        • (g(f(x)))’ = 1

Slide 18

  • Example:
    • Let f(x) = sin(x) and g(x) = arcsin(x) be inverse functions.
    • Find the derivative of (g(f(x)))’ at x = 0.
    • Solution:
      • First, find the composition of the two functions:
        • g(f(x)) = arcsin(sin(x)) = x
      • Now, differentiate both sides with respect to x:
        • (g(f(x)))’ = 1

Slide 19

  • Example:
    • Let f(x) = x^3 and g(x) = cbrt(x) be inverse functions.
    • Find the derivative of (g(f(x)))’ at x = 2.
    • Solution:
      • First, find the composition of the two functions:
        • g(f(x)) = cbrt(x^3) = x
      • Now, differentiate both sides with respect to x:
        • (g(f(x)))’ = 1

Slide 20

  • Summary:
    • Derivatives of inverse functions can be found using the formula: f^(-1)’(x) = 1 / f’(f^(-1)(x))
    • The chain rule for inverse functions allows us to find derivatives of compositions involving inverse functions.
    • Examples showed that the derivatives of compositions of inverse functions simplify to 1.
    • Understanding the derivatives of inverse functions is important for solving various mathematical problems.

Slide 21

  • Applications of Derivatives of Inverse Functions
    • Finding the slope of the tangent line to a curve at a specific point
    • Solving optimization problems
    • Analyzing the behavior of functions

Slide 22

  • Tangent Line to a Curve
    • The derivative of the inverse function represents the slope of the tangent line to the original function at a particular point.
    • Example: Find the equation of the tangent line to the curve y = f(x) = 2x^2 - 3x + 1 at the point (2, f(2)).
    • Solution:
      • Find the derivative of f(x): f’(x) = 4x - 3.
      • Find the inverse function of f(x): y = f(x) = 2x^2 - 3x + 1, x = f^(-1)(y).
      • Find f^(-1)’(y) using the derivative formula: f^(-1)’(y) = 1 / f’(x) = 1 / (4x - 3).
      • Evaluate f^(-1)’(y) at x = 2: f^(-1)’(f(2)) = 1 / (4(2) - 3) = 1 / 5.
      • The slope of the tangent line is 1/5, so the equation is y - f(2) = 1/5(x - 2).

Slide 23

  • Optimization Problems
    • Derivatives of inverse functions can be used to solve optimization problems.
    • Example: A company wants to minimize the cost function C(x) = x^2 + 5x + 10 for producing x units of a product. Find the optimum production level.
    • Solution:
      • Find the derivative of C(x): C’(x) = 2x + 5.
      • Find the inverse function of C(x): y = C(x) = x^2 + 5x + 10, x = C^(-1)(y).
      • Find C^(-1)’(y) using the derivative formula: C^(-1)’(y) = 1 / C’(x) = 1 / (2x + 5).
      • Set C^(-1)’(y) equal to 0 to find the critical point: 1 / (2x + 5) = 0, x = -5/2.
      • The optimum production level is x = -5/2.

Slide 24

  • Behavior of Functions
    • Derivatives of inverse functions provide insights into the behavior of functions.
    • Example: Find the intervals on which the function f(x) = 3x^2 - 2x is increasing or decreasing.
    • Solution:
      • Find the derivative of f(x): f’(x) = 6x - 2.
      • Find the inverse function of f(x): y = f(x) = 3x^2 - 2x, x = f^(-1)(y).
      • Find f^(-1)’(y) using the derivative formula: f^(-1)’(y) = 1 / f’(x) = 1 / (6x - 2).
      • Analyze the signs of f^(-1)’(y):
        • When 6x - 2 > 0, f^(-1)’(y) > 0, which indicates that f(x) is increasing.
        • When 6x - 2 < 0, f^(-1)’(y) < 0, which indicates that f(x) is decreasing.
      • Therefore, f(x) is increasing for x > 1/3 and decreasing for x < 1/3.

Slide 25

  • Summary:
    • Applications of derivatives of inverse functions include finding the slope of tangent lines, solving optimization problems, and analyzing function behavior.
    • Tangent lines to curves can be found using the derivatives of inverse functions.
    • Optimization problems involve finding the minimum or maximum value of a function, which can be done using derivatives of inverse functions.
    • Understanding the behavior of functions is important for various mathematical applications.

Slide 26

  • Recap: Derivatives of Inverse Functions
    • Derivative formula: f^(-1)’(x) = 1 / f’(f^(-1)(x))
    • Chain rule for inverse functions: (f(f^(-1)(x)))’ = f’(f^(-1)(x)) * (f^(-1))’(x)
    • Applications: finding tangent lines, solving optimization problems, analyzing function behavior

Slide 27

  • Key Concepts to Remember
    • Inverse functions are obtained by switching the input and output of a function.
    • Derivatives of inverse functions can be found using the formula f^(-1)’(x) = 1 / f’(f^(-1)(x)).
    • The chain rule for inverse functions allows us to differentiate compositions involving inverse functions.
    • Applications of derivatives of inverse functions include finding tangent lines, solving optimization problems, and analyzing function behavior.

Slide 28

  • Practice Problems
    1. Find the derivative of f^(-1)(x) if f(x) = 4x + 2.
    2. Find the equation of the tangent line to the curve y = f(x) = x^3 - 2x + 1 at the point (1, f(1)).
    3. A company wants to maximize the revenue function R(x) = 2x^2 - 5x + 3 for selling x units of a product. Find the optimum sales level.
    4. Determine the intervals on which the function f(x) = 2x^3 - 6x^2 + 4x is increasing or decreasing.

Slide 29

  • Solution to Practice Problems
    1. Let f(x) = 4x + 2. Find the derivative of f^(-1)(x).
    2. Let f(x) = x^3 - 2x + 1. Find the equation of the tangent line at x = 1.
    3. Let R(x) = 2x^2 - 5x + 3. Find the optimum sales level.
    4. Let f(x) = 2x^3 - 6x^2 + 4x. Determine the intervals of increase and decrease.

Slide 30

  • Conclusion
    • Derivatives of inverse functions are important in understanding the behavior of functions and solving various mathematical problems.
    • Mastery of the concept is essential for success in calculus and related subjects.
    • Further practice and exploration of applications will deepen your understanding and skill in working with inverse functions.