Problem 5 - Calculus for Biology and Medicine

Question

Find the inverse of each function and differentiate each inverse in two ways: (i) Differentiate the inverse function directly, and (ii) use (4.14) to find the derivative of the inverse. $$ f(x)=3-2 x^{3}, x \geq 0 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \), and its derivative is \( -\frac{1}{6}(\frac{3-x}{2})^{-2/3} \).

Step 1: Find the Inverse Function

To find the inverse of the function, start by replacing \( f(x) \) with \( y \): \( y = 3 - 2x^3 \). Solve for \( x \) in terms of \( y \). Re-arranging gives \( 3 - y = 2x^3 \). Divide both sides by 2 to get: \( x^3 = \frac{3 - y}{2} \). Taking the cube root of both sides results in \( x = \sqrt[3]{\frac{3-y}{2}} \). Swap \( x \) and \( y \) to write the inverse: \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \).

Step 2: Differentiate the Inverse Function Directly

The inverse function is \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \). Differentiate it with respect to \( x \) using the chain rule. The derivative of \( \sqrt[3]{u} \) with respect to \( u \) is \( \frac{1}{3}u^{-2/3} \). Let \( u = \frac{3-x}{2} \), then \( \frac{du}{dx} = -\frac{1}{2} \). Therefore, \( \frac{d}{dx}f^{-1}(x) = \frac{1}{3}(\frac{3-x}{2})^{-2/3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{6}(\frac{3-x}{2})^{-2/3} \).

Step 3: Use the Derivative Formula (4.14) to Differentiate

Formula (4.14) states that \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \). The original function is \( f(x) = 3 - 2x^3 \), so \( f'(x) = -6x^2 \). Therefore, \( f'(f^{-1}(x)) = -6[f^{-1}(x)]^2 \). Given that \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \), substitute \( f^{-1}(x) \) to find \( f'(f^{-1}(x)) = -6\left(\sqrt[3]{\frac{3-x}{2}}\right)^2 \). Thus, \( (f^{-1})'(x) = \frac{1}{-6\left(\sqrt[3]{\frac{3-x}{2}}\right)^2} \).

Step 4: Verify the Calculations

Compare the derivatives obtained from both methods: From Step 2, \( -\frac{1}{6}(\frac{3-x}{2})^{-2/3} \) and from Step 3, \( \frac{1}{-6\left(\sqrt[3]{\frac{3-x}{2}}\right)^2} \). Simplifying \( \left(\sqrt[3]{\frac{3-x}{2}}\right)^2 \) is equivalent to \( \left(\frac{3-x}{2}\right)^{2/3} \), thus both derivatives are equivalent, verifying the correctness.

Key Concepts

Differentiation

Differentiation serves as the backbone of calculus and is all about measuring how a function changes as its input changes. Think of it as finding the rate at which something is happening. For instance, how does the position of a car change with time? That's a differentiation at work. In calculus, when we differentiate a function, we typically find the derivative, which is a function itself representing the rate of change.

• For any function like \( f(x) = 3 - 2x^3 \), differentiating it means finding \( f'(x) \), the derivative.
• This derivative tells us how \( f(x) \) changes in response to changes in \( x \).

In the original exercise, differentiation helped simplify the process of finding the inverse's derivative. By using it, we focus on how the output of the function plugs into the input of its inverse, covering all changes effectively.

Chain Rule

The chain rule is a fundamental tool in calculus used when differentiating composite functions. A composite function is just one function stuffed into another. It might sound complicated, but it's like peeling an onion, dealing with each layer one by one.

• Consider the inverse function \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \).
• It is a composite because it's made of: a cube root and a linear subtraction.

To differentiate \( f^{-1}(x) \) using the chain rule, figure out the derivative of the outer function, while treating the inner function as a single unit. Then, multiply by the derivative of that inner unit.

• For the outer function, if \( g(u) = u^{1/3} \), its derivative is \( \frac{1}{3}u^{-2/3} \).
• The inner function here, \( u = \frac{3-x}{2} \), has a derivative of \( -\frac{1}{2} \).
• Putting it all together, applying the chain rule gets us to the correct derivative step by step.

Derivative Formula

The derivative formula used in this context is represented by a nifty trick—if you have an inverse function, an easier way to differentiate is to use the formula: \[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \]. This formula requires two ingredients:

• The derivative of the original function \( f(x) \), like \( f'(x) = -6x^2 \).
• The value of the function, \( f^{-1}(x) \), fed back through \( f(x) \).

Knowing \( f^{-1}(x) = \sqrt[3]{\frac{3-x}{2}} \), we substitute as:\\(-6(\sqrt[3]{\frac{3-x}{2}})^2 \) which translates to \(\left(\frac{3-x}{2}\right)^{2/3}\). This computation checks if our prior results align when applying this method. It's a neat and tidy shortcut to verify your inverse derivative.