Problem 14 - 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 \(g^{\prime}\). $$ h(x)=\sqrt{f(x) g(x)} $$

Step-by-Step Solution

Answer

The derivative \(h'(x)\) of the function \(h(x)=\sqrt{f(x) g(x)}\) is given by \(h'(x) = \frac{1}{2} (f(x)g(x))^{-1/2} * (f'(x)g(x) + f(x)g'(x)) \).

Step 1: Rewrite the square root function

First, rewrite the square root function \(h(x)=\sqrt{f(x) g(x)}\) as \(h(x) = [f(x) g(x)]^{1/2}\). This step simplifies the differentiation process, because now we have a power function, which is easier to differentiate.

Step 2: Use the chain rule for differentiation

Start by applying the chain rule for differentiation, which states that the derivative of a function of a function is the derivative of the outer function times the derivative of the inner function. The outer function is \(h(x)=(u)^{1/2}\) and the inner function is \(u=f(x) g(x)\). Apply the chain rule: \(h'(x) = \frac{1}{2} (u)^{-1/2} * u'\).

Step 3: Use the product rule for differentiation

Now use the product rule to find the derivative of the inner function \(u = f(x) g(x)\). The product rule for differentiation states that the derivative of two multiplied functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. So, we have \(u' = f'(x)g(x) + g'(x)f(x)\). Substituting this into the equation from step 2 gives \(h'(x) = \frac{1}{2} (f(x)g(x))^{-1/2} * (f'(x)g(x) + f(x)g'(x)) \).

Step 4: Simplify if necessary

Finally, simplify the equation if necessary. This can involve factoring or reducing terms. In this case, no further simplification is necessary once the product rule has been correctly applied.

Key Concepts

The Chain Rule

The chain rule is a fundamental concept in calculus, particularly useful when dealing with composite functions—functions inside other functions. If you need to find the derivative of a composition of functions, the chain rule offers a straightforward strategy.
In essence, this rule helps break down complex derivative problems into manageable pieces.

• To apply the chain rule, first identify the outer and inner functions in your composite function.
• The derivative of the composite function is found by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function itself.

For example, in the problem above, you identified the outer function as a square root, rewritten as a power function \(h(x) = [u]^{1/2}\), where \(u = f(x)g(x)\) is the inner function.
So the chain rule says \(h'(x) = \frac{1}{2} [u]^{-1/2} \cdot u'\), where \(u'\) will be found using additional differentiation rules.
Using the chain rule effectively simplifies the process of differentiating complicated compositions by systematically evaluating and combining derivatives.

The Product Rule

The product rule is a valuable tool in calculus for differentiating products of functions. It states that the derivative of the product of two functions is not merely the product of their derivatives.
Instead, it takes into account the change in each function relative to the other.

• When you have a function \(u(x) = f(x)g(x)\) composed of two differentiable functions, the derivative is found by applying the product rule.
• The rule is written as: \(u'(x) = f'(x)g(x) + g'(x)f(x)\).

In the solution process for the given function, after applying the chain rule, the product rule is applied to the inner function \(u = f(x)g(x)\).
This allows us to find \(u'(x) = f'(x)g(x) + g'(x)f(x)\) and substitute it back into the equation derived from the chain rule.
The product rule efficiently separates the differentiation process into two straightforward calculations, making it essential for accurately handling products in calculus.

Differentiable Functions

Understanding differentiable functions is crucial to applying calculus concepts effectively, such as the chain and product rules. A function is called differentiable at a point if it has a derivative there; more broadly, it is differentiable over a range if it has derivatives throughout that range.

Differentiability ensures:

• The function is continuous at the point of interest.
• There are no sharp corners or discontinuities.

In the given problem, we assume that \(f\) and \(g\) are differentiable everywhere, which guarantees that the operations needed for finding \(h'(x)\) using the chain and product rules are valid.
This assumption is important as it ensures the derivative processes applied to both functions \(f\) and \(g\) are legitimate. Without differentiability, derivatives cannot be accurately calculated, undermining the entire methodology of finding \(h'(x)\).

Therefore, understanding and confirming differentiability is a foundational step for tackling calculus problems involving these derivative rules.