Problem 35 - Calculus for Biology and Medicine

Question

Suppose that \(f^{\prime}(x)=\frac{1}{x}\). Find the following: (a) \(\frac{d}{d x} f\left(x^{2}+3\right)\) (b) \(\frac{d}{d x} f(\sqrt{x-1})\)

Step-by-Step Solution

Answer

(a) \( \frac{2x}{x^2 + 3} \); (b) \( \frac{1}{2(x-1)} \)."

Step 1: Understand the Problem

We are given that \( f'(x) = \frac{1}{x} \). We need to find the derivatives \( \frac{d}{d x} f\left(x^{2}+3\right) \) and \( \frac{d}{d x} f(\sqrt{x-1}) \). First, we recognize that these require the application of the chain rule in differentiation.

Step 2: Differentiate Part (a) Using Chain Rule

For (a), we need to find \( \frac{d}{dx} f(x^2 + 3) \). By the chain rule, if \( g(x) = x^2 + 3 \), then \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). Here, \( f'(g(x)) = \frac{1}{g(x)} \) and \( g'(x) = 2x \). Thus, the derivative is \[ \frac{d}{dx} f(x^2 + 3) = \frac{1}{x^2 + 3} \cdot 2x = \frac{2x}{x^2 + 3}. \]

Step 3: Differentiate Part (b) Using Chain Rule

For (b), we need \( \frac{d}{dx} f(\sqrt{x-1}) \). Let \( h(x) = \sqrt{x-1} = (x-1)^{1/2} \). By the chain rule, \( \frac{d}{dx} f(h(x)) = f'(h(x)) \cdot h'(x) \). Here, \( f'(h(x)) = \frac{1}{h(x)} \) and \( h'(x) = \frac{1}{2}(x-1)^{-1/2} \). Thus, the derivative is \[ \frac{d}{dx} f(\sqrt{x-1}) = \frac{1}{\sqrt{x-1}} \cdot \frac{1}{2\sqrt{x-1}} = \frac{1}{2(x-1)}. \]

Key Concepts

Chain Rule

The chain rule is a powerful technique in calculus. It's used to differentiate composite functions. When you have a function within another function, like \( f(g(x)) \), the chain rule becomes handy. This rule states that to find the derivative of \( f(g(x)) \), you multiply the derivative of the outer function \( f' \) by the derivative of the inner function \( g' \).

Here's the formula expressed in more detail:

• Let \( y = f(u) \), where \( u = g(x) \).
• Then \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).

To understand better, think of peeling an onion. You have to peel each layer one by one. Similarly, in the chain rule, you differentiate each layer starting from the outermost to the innermost. Each "layer" is a separate function that you work with.

Derivatives

Derivatives tell us the rate at which a function is changing. It’s like watching the speedometer of a car to understand how fast it’s going at any moment.

To calculate a derivative, you typically use rules based on the form of the function. The common rules include:

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Product Rule: For functions \( u(x) \) and \( v(x) \), \( (uv)' = u'v + uv' \).
• Quotient Rule: For \( u(x) \) and \( v(x) \), \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
• Chain Rule: Already explained above.

Derivatives are fundamental in calculus because they help in understanding instantaneous change. For example, with \( f'(x) = \frac{1}{x} \), it shows how the function \( f \) changes as \( x \) changes. The exercises we solved use these principles to find the derivatives of complex functions.

Function Notation

Function notation is a way to express mathematical relationships. Think of it like a machine: you input a number, and the machine (function) transforms it into another number.

In function notation, \( f(x) \) indicates a function named \( f \) with \( x \) as the input variable.

• For example, if \( f(x) = x^2 + 3 \), then \( f(x) \) tells you what to do with \( x \), specifically to square \( x \) and add 3.
• The notation \( f(g(x)) \) means you first apply \( g(x) \) to \( x \), then apply \( f(x) \) to the result.
• This is critical when using the chain rule because you need to understand which function operates inside another.

Understanding how to use function notation is crucial in calculus. It provides clarity and structure when working with complex math problems, especially when differentiating functions.