Finding The Derivative Of An Inverse Function

Unraveling the derivative of an inverse function might seem like navigating a mathematical labyrinth at first glance. However, with a systematic approach and a firm grasp of the underlying principles, you can confidently tackle this fascinating area of calculus. This comprehensive guide will walk you through the process, providing clear explanations, step-by-step examples, and practical insights to help you master the derivative of inverse functions.

Understanding Inverse Functions

Before diving into the derivative, let's solidify our understanding of inverse functions. An inverse function, denoted as f⁻¹(x), essentially "undoes" the original function f(x). More formally, if y = f(x), then x = f⁻¹(y).

Key Properties of Inverse Functions:

• Reflection: The graph of an inverse function is a reflection of the original function across the line y = x.
• Domain and Range: The domain of f(x) is the range of f⁻¹(x), and vice versa.
• Composition: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for all x in their respective domains.
• One-to-one: A function must be one-to-one (pass the horizontal line test) to have an inverse function. This means that for every y value, there is only one corresponding x value.

Example:

Consider the function f(x) = 2x + 3. To find its inverse, we can follow these steps:

1. Replace f(x) with y: y = 2x + 3
2. Swap x and y: x = 2y + 3
3. Solve for y:
   • x - 3 = 2y
   • y = (x - 3) / 2
4. Replace y with f⁻¹(x): f⁻¹(x) = (x - 3) / 2

Therefore, the inverse function of f(x) = 2x + 3 is f⁻¹(x) = (x - 3) / 2.

The Derivative of an Inverse Function: The Formula

The core concept we need is the formula for finding the derivative of an inverse function. This formula allows us to calculate the derivative of f⁻¹(x) at a specific point without explicitly finding the inverse function itself.

The Formula:

If f(x) is a differentiable function with an inverse function f⁻¹(x), and f'(f⁻¹(x)) ≠ 0, then the derivative of the inverse function is given by:

(f⁻¹)'(x) = 1 / f'(f⁻¹(x))

Explanation:

• (f⁻¹)'(x) represents the derivative of the inverse function with respect to x.
• f'(x) represents the derivative of the original function with respect to x.
• f'(f⁻¹(x)) means we are evaluating the derivative of the original function at the point f⁻¹(x).

In simpler terms: The derivative of the inverse function at a point x is the reciprocal of the derivative of the original function evaluated at the inverse function of x.

Steps to Find the Derivative of an Inverse Function

Here's a detailed step-by-step guide to finding the derivative of an inverse function:

1. Verify the Existence of the Inverse Function: Ensure that the original function f(x) is one-to-one over the interval of interest. You can do this graphically using the horizontal line test or algebraically by showing that f(a) = f(b) implies a = b.

2. Find f'(x): Calculate the derivative of the original function f(x). This is a standard differentiation process using the rules of calculus (power rule, product rule, quotient rule, chain rule, etc.).

3. Find f⁻¹(x) or f⁻¹(a) (if necessary): Sometimes, you need to find the explicit form of the inverse function f⁻¹(x). Other times, you'll be asked to find the derivative of the inverse at a specific point x = a, in which case you'll need to find f⁻¹(a). If finding the explicit form of f⁻¹(x) is difficult, you can often proceed without it (see step 5).

4. Evaluate f'(f⁻¹(x)) or f'(f⁻¹(a)): Substitute the inverse function f⁻¹(x) (or the value f⁻¹(a)) into the derivative of the original function f'(x). This gives you f'(f⁻¹(x)) (or f'(f⁻¹(a))).

5. Apply the Formula: Use the formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)) (or (f⁻¹)'(a) = 1 / f'(f⁻¹(a)))) to find the derivative of the inverse function.

Examples with Detailed Solutions

Let's work through some examples to illustrate the process.

Example 1:

Given f(x) = x³ + 2x - 1, find (f⁻¹)'(2).

1. Verify Existence: f(x) = x³ + 2x - 1 is a monotonically increasing function (its derivative is always positive), so it is one-to-one and has an inverse.

2. Find f'(x): f'(x) = 3x² + 2

3. Find f⁻¹(2): We need to find the value of x such that f(x) = 2. So, we need to solve the equation: x³ + 2x - 1 = 2 which simplifies to x³ + 2x - 3 = 0. By observation or using the rational root theorem, we find that x = 1 is a root. Therefore, f⁻¹(2) = 1.

4. Evaluate f'(f⁻¹(2)): f'(f⁻¹(2)) = f'(1) = 3(1)² + 2 = 5

5. Apply the Formula: (f⁻¹)'(2) = 1 / f'(f⁻¹(2)) = 1 / 5

Therefore, (f⁻¹)'(2) = 1/5.

Example 2:

Given f(x) = √(x + 1) for x ≥ -1, find (f⁻¹)'(x).

1. Verify Existence: f(x) = √(x + 1) is one-to-one for x ≥ -1 since it is a strictly increasing function.

2. Find f'(x): f(x) = (x + 1)^(1/2). Using the power rule and chain rule: f'(x) = (1/2)(x + 1)^(-1/2) = 1 / (2√(x + 1))

3. Find f⁻¹(x):

   • y = √(x + 1)
   • x = √(y + 1)
   • x² = y + 1
   • y = x² - 1
   • f⁻¹(x) = x² - 1 for x ≥ 0 (since the range of f(x) is y ≥ 0)

4. Evaluate f'(f⁻¹(x)): f'(f⁻¹(x)) = f'(x² - 1) = 1 / (2√((x² - 1) + 1)) = 1 / (2√(x²)) = 1 / (2|x|) = 1 / (2x) (since x ≥ 0)

5. Apply the Formula: (f⁻¹)'(x) = 1 / f'(f⁻¹(x)) = 1 / (1 / (2x)) = 2x

Therefore, (f⁻¹)'(x) = 2x for x ≥ 0.

Example 3:

Let f(x) = x⁵ + x + 1. Find the derivative of the inverse function at x = 3.

1. Verify Existence: The derivative of f(x) is f'(x) = 5x⁴ + 1. Since f'(x) is always positive, f(x) is strictly increasing and therefore one-to-one, guaranteeing the existence of an inverse function.

2. Find f'(x): As we found in step 1, f'(x) = 5x⁴ + 1.

3. Find f⁻¹(3): We need to find the value of x such that f(x) = 3. This means solving x⁵ + x + 1 = 3, which simplifies to x⁵ + x - 2 = 0. By observation, we can see that x = 1 is a solution, since 1⁵ + 1 - 2 = 0. Therefore, f⁻¹(3) = 1.

4. Evaluate f'(f⁻¹(3)): f'(f⁻¹(3)) = f'(1) = 5(1)⁴ + 1 = 6.

5. Apply the Formula: (f⁻¹)'(3) = 1 / f'(f⁻¹(3)) = 1 / 6.

Therefore, the derivative of the inverse function at x = 3 is 1/6.

Example 4: Trigonometric Functions

Let f(x) = sin(x), where -π/2 ≤ x ≤ π/2. Find (f⁻¹)'(1/2).

1. Verify Existence: On the interval -π/2 ≤ x ≤ π/2, sin(x) is strictly increasing and therefore one-to-one, so its inverse exists. The inverse function is arcsin(x) or sin⁻¹(x).

2. Find f'(x): f'(x) = cos(x)

3. Find f⁻¹(1/2): We need to find the value of x such that sin(x) = 1/2. On the interval -π/2 ≤ x ≤ π/2, this occurs at x = π/6. Therefore, f⁻¹(1/2) = π/6.

4. Evaluate f'(f⁻¹(1/2)): f'(f⁻¹(1/2)) = f'(π/6) = cos(π/6) = √3 / 2

5. Apply the Formula: (f⁻¹)'(1/2) = 1 / f'(f⁻¹(1/2)) = 1 / (√3 / 2) = 2 / √3 = (2√3) / 3

Therefore, (f⁻¹)'(1/2) = (2√3) / 3.

Common Mistakes to Avoid

• Forgetting to verify the existence of the inverse function: Not all functions have inverses. Make sure the function is one-to-one before proceeding.
• Incorrectly calculating f'(x): A mistake in finding the derivative of the original function will propagate through the entire process. Double-check your differentiation.
• Confusing f⁻¹(x) with 1/f(x): The inverse function is not the same as the reciprocal of the function.
• Incorrectly evaluating f'(f⁻¹(x)): Ensure you substitute the entire inverse function into the derivative, not just x.
• Forgetting the reciprocal: The final step involves taking the reciprocal of f'(f⁻¹(x)). Don't forget this crucial step!
• Domain Issues: Be mindful of the domains of both the original function and its inverse. The derivative of the inverse is only valid where the inverse function is differentiable.

Applications of the Derivative of Inverse Functions

The derivative of inverse functions has applications in various areas of mathematics and science:

• Related Rates: It can be used to solve related rates problems where the rate of change of an inverse relationship is needed.
• Optimization: It can help find the maximum or minimum values of functions involving inverse relationships.
• Physics: It can be used in physics to analyze inverse relationships between physical quantities.
• Engineering: It can be applied in engineering problems involving inverse functions.
• Economics: Derivatives of inverse functions can be used to analyze supply and demand curves.

A Deeper Dive into the Theory

The formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)) can be derived using the chain rule. Recall that f(f⁻¹(x)) = x. Differentiating both sides with respect to x using the chain rule, we get:

f'(f⁻¹(x)) * (f⁻¹)'(x) = 1

Solving for (f⁻¹)'(x), we obtain:

(f⁻¹)'(x) = 1 / f'(f⁻¹(x))

This derivation highlights the fundamental connection between the derivative of a function and the derivative of its inverse.

Advanced Techniques and Special Cases

While the formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)) is the cornerstone, some situations require advanced techniques:

• Implicit Differentiation: When f(x) is defined implicitly, you may need to use implicit differentiation to find f'(x).
• Parametric Equations: If f(x) is defined parametrically, you'll need to use parametric differentiation techniques.
• Multivariable Calculus: The concept extends to multivariable calculus, where you'll deal with partial derivatives and the inverse function theorem.

Practice Problems

To solidify your understanding, try these practice problems:

1. If f(x) = x³ + 1, find (f⁻¹)'(9).
2. If f(x) = e^(2x), find (f⁻¹)'(x).
3. Let f(x) = x + cos(x). Find the derivative of the inverse function at x = 1. (Hint: f(0) = 1)
4. If f(x) = tan(x) for -π/2 < x < π/2, find (f⁻¹)'(1).
5. Given f(x) = x⁵ + 3x³ + x, find (f⁻¹)'(5). (Hint: What value of x makes f(x) = 5?)

Conclusion

Finding the derivative of an inverse function is a valuable skill in calculus. By understanding the relationship between a function and its inverse, mastering the derivative formula, and practicing with examples, you can confidently solve a wide range of problems. Remember to verify the existence of the inverse, carefully calculate the derivatives, and avoid common mistakes. With consistent practice, you'll be able to navigate the world of inverse functions and their derivatives with ease. The key is to break down the problem into manageable steps and apply the formula methodically. Happy differentiating!