Problem 18 - 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 \(g^{\prime}\). $$ h(x)=[f(x)]^{3} g(2 x) $$

Step-by-Step Solution

Answer

\(h'(x) = 3[f(x)]^2 f'(x) g(2x) + 2 [f(x)]^3 g'(2x)\)

Step 1: Differentiate the first function

We use the chain rule to differentiate \(f(x)^3\), which results in \(3[f(x)]^2 f'(x)\).

Step 2: Differentiate the second function

We use the chain rule to differentiate \(g(2x)\), which results in \(g'(2x) * 2\).

Step 3: Apply the product rule

We use the product rule to differentiate \(h(x)\). This rule is \((u*v)'=u'*v + u*v'\). Applying this, we get \(h'(x) = [3[f(x)]^2 f'(x)]*g(2x) + [f(x)]^3*[2g'(2x)]\).

Key Concepts

Chain Rule

When faced with the task of differentiating composite functions, the chain rule is an indispensable tool. It allows us to take derivatives of functions that are nested within other functions. It can be thought of as a mathematical 'unpacking' process. For example, to differentiate a function like \( h(x) = [f(x)]^3 \), the chain rule instructs us to first differentiate the outer function, which in this case is \( u^3 \) with \( u=f(x) \) and then multiply by the derivative of the inner function, \( f(x) \). This results in the derivative \( h'(x) = 3[f(x)]^2 f'(x) \) as the chain rule implies performing \( d/dx(u^3) \cdot du/dx \).

Apply the chain rule especially when you have functions 'within' functions, and differentiate from the outside in. When you see an expression like \( g(2x) \), recognize that \( 2x \) is the inner function and \( g \) the outer function. The derivative, therefore, involves taking the derivative of \( g \) with respect to its argument—denoted as \( g'(2x) \)—and then multiplying it by the derivative of the inner function, which is \( 2 \). This process systematically tackles complex differentiation and ensures no part of the function is overlooked.

Product Rule

At times, functions are not nested but rather multiplied with one another. This is where the product rule comes into play. It states that if you have two functions, say \( u(x) \) and \( v(x) \) that are multiplied together to form \( h(x) = u(x) \)\times \ v(x), the derivative of \( h(x) \) is not simply the derivative of \( u \) times the derivative of \( v \) but rather \( h'(x) = u'(x)v(x) + u(x)v'(x) \).

For our example, \( h(x)=[f(x)]^{3} g(2 x) \) consists of \( u(x)=[f(x)]^3 \) and \( v(x)=g(2x) \). Applying the product rule, we differentiate \( u(x) \) and \( v(x) \) separately, then combine them as \( h'(x) = u'(x)v(x) + u(x)v'(x) \) which gives us \( h'(x) = [3[f(x)]^2 f'(x)]*g(2x) + [f(x)]^3*[2g'(2x)] \). This formula is crucial for finding the derivative of products of functions and serves as a key strategy in calculus for deciphering more complex equations.

Derivatives of Functions

Derivatives represent the rate at which a function is changing at any point and are foundational in calculus. Understanding the derivative of a function means understanding how sensitive a function is to change in its input. For example, if \( f(x) \) represents position over time, \( f'(x) \) represents velocity—the rate of change of position.

To find a derivative effectively, you must be familiar with the basic rules of differentiation, such as the power rule, where the derivative of \( x^n \) is \( nx^{n-1} \) and general formulas for common functions like exponential and trigonometric functions. It's important to start with these fundamental rules before attempting more complex applications such as the chain rule and product rule. Remember, practice is key, and using proper notation will prevent confusion, especially when dealing with higher-order derivatives or when applying the rules in combination, as is often required in calculus exercises.