Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 13 through 18, find \(h^{\prime}(x) .\) Assume that \(f\) and \(g\) are differentiable on \((-\infty, \infty)\). Your answers may be in terms of \(f, g, f^{\prime}\), and \(g^{\prime}\). $$ h(x)=f\left(x^{2}\right) e^{g(x)} $$

Step-by-Step Solution

Answer

The derivative of the function \(h(x)\) is given by \(h^{\prime}(x) = f^{\prime}(x^{2} ) \cdot 2x \cdot e^{g(x)} + f\left(x^{2}\right) \cdot e^{g(x)} \cdot g^{\prime}(x)\).

Step 1: Identify the Inner and Outer Functions

Take a look at the given function \(h(x)\). We have two 'parts' here that are multiplied together: \(f\left(x^{2}\right)\) and \(e^{g(x)}\). For each of these two parts, we need to identify the inner function (function of x) and outer function.

Step 2: Apply the Product Rule

Start by applying the product rule, which states that the derivative of two functions multiplied together is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Step 3: Apply the Chain Rule to Each Part

Next, apply the chain rule to each part. The chain rule states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. Thus, we have for the first part, \(f^{\prime}(x^{2})\cdot2x\) (derivative of \(f(x^{2})\)) and for the second part, \( e^{g(x)} \cdot g^{\prime}(x)\) (derivative of \( e^{g(x)}\)).

Step 4: Combine and Simplify

Finally, combine the results from applying the product rule and the chain rule and simplify.

Key Concepts

Product Rule

When faced with the task of differentiating the function given in the exercise, one of the core concepts we rely on is the product rule. The product rule is essential when we need to find the derivative of two or more functions multiplied together. In simpler terms, if we have functions \( u(x) \) and \( v(x) \) and they are being multiplied (forming \( h(x) = u(x)v(x) \)), then the derivative \( h'(x) \) is given by:

• \( h'(x) = u'(x)v(x) + u(x)v'(x) \)

Each part of the formula represents how one piece of the product changes, while keeping the other unchanged — a crucial point for solving calculus problems involving products of functions.
It's a direct idea: think of how changing one aspect (function) while the other remains constant affects the result, then simply switch, and add.

Chain Rule

Differentiating composite functions often requires the use of the chain rule — another fundamental in calculus problem solving. This rule simplifies the process when we have functions within functions. For a function \( h(x) = f(g(x)) \), the chain rule tells us:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

This means, first take the derivative of the outer function, keeping the inner function unchanged, then multiply it by the derivative of the inner function.
In applying this to our original exercise, each part involving \( f(x^2) \) and \( e^{g(x)} \) utilizes this principle to capture the way each 'inner' part influences the whole.

Differentiable Functions

A differentiable function is a function for which the derivative exists at each point in its domain. This concept underpins many of the rules in calculus, such as the chain and product rules. In our exercise, knowing that \( f \) and \( g \) are differentiable is crucial — it allows us to confidently apply these rules.

• If \( f \) and \( g \) weren’t differentiable, taking derivatives wouldn’t be straightforward, limiting our ability to analyze changes.

It's more than a technical term; it’s a guarantee that the functions behave nicely, allowing calculus methods to be effective. Each smooth curve or line that represents such functions opens the door for easier mathematical analysis and manipulation.

Composite Function

A composite function involves the combination of two functions, where one function is applied to the results of another. The exercise’s function, \( h(x) = f(x^2) e^{g(x)} \), highlights this by involving \( f(x^2) \), a clear example of composition.

• This nesting of functions, where output from one feeds into another's input, is common in calculus problems.

Composite functions are crucial when solving calculus problems since they require special rules, like the chain rule, for differentiation.
Understanding these layers in mathematics allows for a deep comprehension of how functions can interact, particularly in modeling complex systems.

Calculus Problem Solving

Solving calculus problems like this one involves mastering a variety of methods and rules. It’s about efficiently and correctly applying core concepts like the product and chain rules to find derivatives, often working through multi-step processes.

• Identify and analyze the structure of the function before starting.
• Carefully apply each rule step by step.

Problem solving is much like piecing a puzzle together; each small technique or tool contributes to the solution. With practice, these principles become second nature, resulting in clearer and faster solutions.