Problem 34 - Calculus for Biology and Medicine

Question

Suppose that \(f^{\prime}(x)=2 x+1\). Find the following: (a) \(\frac{d}{d x} f\left(x^{2}\right)\) at \(x=-1\) (b) \(\frac{d}{d x} f(\sqrt{x})\) at \(x=4\)

Step-by-Step Solution

Answer

(a) \(-6\), (b) \(\frac{5}{4}\).

Step 1: Understanding the Problem

We are asked to find the derivatives of two compositions of functions, \(f(x^2)\) and \(f(\sqrt{x})\), and evaluate them at specific points using the chain rule.

Step 1: Derivative of a Composite Function (a)

First, apply the chain rule to find the derivative of \(f(x^2)\). The chain rule states that if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Here, \(g(x) = x^2\), so \(g'(x) = 2x\). Thus, \(\frac{d}{dx} f(x^2) = f'(x^2) \cdot 2x\).

Step 2: Evaluate the Derivative at x = -1 (a)

Substitute \(x = -1\) into the derivative expression. \(\frac{d}{dx} f(x^2) = f'((-1)^2) \cdot 2(-1) = f'(1) \cdot (-2)\). Given that \(f'(x) = 2x + 1\), substitute \(x = 1\) to find \(f'(1) = 2(1) + 1 = 3\). Therefore, \(\frac{d}{dx} f(x^2)\) at \(x = -1\) is \(3 \cdot (-2) = -6\).

Step 3: Derivative of a Composite Function (b)

Now, apply the chain rule to find the derivative of \(f(\sqrt{x})\). Here, \(g(x) = \sqrt{x}\), so \(g'(x) = \frac{1}{2\sqrt{x}}\). Thus, \(\frac{d}{dx} f(\sqrt{x}) = f'(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\).

Step 4: Evaluate the Derivative at x = 4 (b)

Substitute \(x = 4\) into the derivative expression. \(\frac{d}{dx} f(\sqrt{x}) = f'(\sqrt{4}) \cdot \frac{1}{2\cdot2} = f'(2) \cdot \frac{1}{4}\). From \(f'(x) = 2x + 1\), \(f'(2) = 2(2) + 1 = 5\). Thus, \(\frac{d}{dx} f(\sqrt{x})\) at \(x = 4\) is \(5 \cdot \frac{1}{4} = \frac{5}{4}\).

Key Concepts

Understanding Derivatives

Derivatives are a core concept in calculus that measure how a function changes as its input changes. In simpler terms, a derivative tells us the slope of a function at any given point. Think of it as the rate of change. For example, when a person walks, their speed can increase or decrease. The rate at which their speed changes is similar to what a derivative represents for a function. In mathematical terms, if we have a function \( f(x) \), its derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). These notations represent the same idea, showing how the output \( f(x) \) changes with respect to small changes in the input \( x \). Understanding derivatives is essential in solving calculus problems, particularly when dealing with more complex expressions as seen in the problem, where functions are nested within other functions.

Composite Functions and Their Derivatives

A composite function is created when one function is applied to the result of another function. Imagine first making coffee, then adding cream. Each step processes the previous step's result – that's a composite function! Mathematically, if \( g(x) \) is our first function and we apply \( f(x) \) to it, we get \( f(g(x)) \). This nesting of functions can make finding derivatives tricky.To solve for the derivative of a composite function, the chain rule is our friend. The chain rule formula states that if \( y = f(g(x)) \), the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

• Example 1: For \( f(x^2) \), \( g(x) = x^2 \). The derivative \( g'(x) = 2x \).
• Example 2: For \( f(\sqrt{x}) \), \( g(x) = \sqrt{x} \). The derivative \( g'(x) = \frac{1}{2\sqrt{x}} \).

Using the chain rule helps us break down the problem and find solutions for "nested" functions.

Step-by-Step Calculus Problem Solving

Solving calculus problems, especially those involving chain rule applications or composite functions, is all about understanding and applying the right steps repeatedly. Each step builds on the previous one. Here is a simplified approach to tackle such problems:

• Identify and Understand: First, understand what the problem is asking you to find. Look at what functions are involved and if they are composite.
• Apply the Chain Rule: Clearly identify the outer function \( f \) and the inner function \( g \). Then use the formula \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \) to find the derivative.
• Substitute Values: Once the derivatives have been calculated, substitute the specific values the problem asks for. This gives you the final answer.

Practice these steps with different functions to become more familiar and confident in solving calculus problems. Calculus involves a lot of practice and understanding, but with time, concepts like the chain rule become intuitive.