Problem 4 - 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 (4.12) to find the derivative of the inverse. $$ f(x)=3 x^{2}+2, x \geq 0 $$

Step-by-Step Solution

Answer

The inverse of the function is \(f^{-1}(y) = \sqrt{\frac{y - 2}{3}}\). Its derivative is \(\frac{1}{\sqrt{3(y - 2)}}\) by both methods.

Step 1: Find the Inverse Function

First, you need to express the given function in terms of \(y\). Set \(y = f(x) = 3x^2 + 2\). Solve for \(x\) in terms of \(y\):\[ y = 3x^2 + 2 \Rightarrow x^2 = \frac{y - 2}{3} \Rightarrow x = \sqrt{\frac{y - 2}{3}}, \,\text{since}\, x \geq 0.\]Thus, the inverse function is \(f^{-1}(y) = \sqrt{\frac{y - 2}{3}}\)."},{

Step 2: Differentiate the Inverse Function Directly

Differentiate the inverse function directly with respect to \(y\).The inverse function is \(f^{-1}(y) = \sqrt{\frac{y - 2}{3}}\).Use the chain rule to differentiate:\[ \frac{d}{dy}(f^{-1}(y)) = \frac{1}{2}\left(\frac{1}{3}(y - 2)\right)^{-1/2} \cdot \frac{1}{3} = \frac{1}{2 \sqrt{\frac{y - 2}{3}}} \cdot \frac{1}{3}.\]Simplify this to get:\[ \frac{d}{dy}(f^{-1}(y)) = \frac{1}{2 \cdot 3 \cdot \sqrt{\frac{y - 2}{3}}} = \frac{1}{3 \sqrt{\frac{y - 2}{3}}} = \frac{1}{\sqrt{3(y - 2)}}.\]"},{

Step 3: Use Formula (4.12) to Differentiate the Inverse

Using formula (4.12), which is \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))},\) first compute the derivative of \(f(x)\):\[ f'(x) = \frac{d}{dx}(3x^2 + 2) = 6x.\]Then, substitute \(x = f^{-1}(y) = \sqrt{\frac{y - 2}{3}}\) into \(f'(x)\):\[ f'(\sqrt{\frac{y - 2}{3}}) = 6\left(\sqrt{\frac{y - 2}{3}}\right).\]Thus, using the Formula (4.12):\[ (f^{-1})'(y) = \frac{1}{6 \sqrt{\frac{y - 2}{3}}}.\]

Key Concepts

Understanding Calculus

Calculus is like the bridge that connects the physical world with mathematics. It allows us to understand how things change and how they are moved. The core activities in calculus are differentiation and integration. Here, we're concerned with differentiation, which is the process of finding a derivative.
- A core idea in calculus is to find how a function changes at any point.
- Differentiation transforms a function into another function that represents its rate of change.
- By using derivatives, we can calculate slopes of tangent lines, maxima and minima of functions, and even solve complex problems in physics and engineering.

In solving problems, and especially when finding derivatives of functions and their inverses, calculus becomes invaluable. This exercise shows us one way calculus helps us understand and manipulate functions analytically, offering insights into problems that appear in real-world applications.

The Role of Derivative

The derivative of a function gives us the rate at which the function's value changes as its input changes. This is an essential concept because:
- It tells us the slope of the tangent line to the function at any point.
- A positive derivative indicates increasing behavior, while a negative derivative shows decreasing behavior.

In the exercise, we first find the derivative of the original function: \(f(x) = 3x^2 + 2\), which is \(f'(x) = 6x\). This tells us how steeply the function rises as \(x\) increases.
For inverse functions, the derivative plays a crucial role in understanding how the inverse function behaves. We used the derivative of the original function to help find the derivative of the inverse function more efficiently using formula (4.12).
Analyzing derivatives shows us how tiny bumps in the input cause changes in the output, allowing for predictions and deeper comprehension of the functions.

Mastering the Chain Rule

The chain rule is a powerful technique used in calculus to differentiate composite functions. Here's why it's useful:
- It allows us to break down complex functions into simpler parts, making differentiation manageable.
- With the chain rule, we can handle functions nested inside one another.

For example, in our exercise, we applied the chain rule to differentiate the inverse function \(f^{-1}(y) = \sqrt{\frac{y - 2}{3}}\). The chain rule helped dissect this function into simpler components to find its derivative effectively.
In simple terms, the chain rule states that if you have a function inside another, first differentiate the outer function, keeping the inner function intact, and then differentiate the inner function. Finally, you multiply these derivatives.
By mastering the chain rule, we enhance our ability to work with complicated functions effortlessly, focusing on each part of the function step by step, simplifying the analysis significantly.