Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated derivative. $$ g^{\prime}\left(\frac{1}{2}\right) \text { if } g(s)=\cos \pi s \sin ^{2} \pi s $$

Step-by-Step Solution

Answer

The derivative is \( g'(\frac{1}{2}) = -\pi \).

Step 1: Differentiate the Function

We start by finding the derivative of the function \( g(s) = \cos(\pi s) \sin^2(\pi s) \). Use the product rule: if \( u = \cos(\pi s) \) and \( v = \sin^2(\pi s) \), then\( g'(s) = u'v + uv' \).

Step 2: Apply the Product Rule

First, find the derivatives of \( u \) and \( v \). For \( u = \cos(\pi s) \), \( u' = -\pi \sin(\pi s) \). For \( v = \sin^2(\pi s) \), use the chain rule: \( v' = 2\sin(\pi s) \cdot \pi \cos(\pi s) = 2\pi \sin(\pi s) \cos(\pi s) \).

Step 3: Combine the Derivative Terms

Substitute back into the product rule formula: \( g'(s) = [-\pi \sin(\pi s)]\sin^2(\pi s) + \cos(\pi s)[2\pi \sin(\pi s) \cos(\pi s)] \). Simplify this to get \( g'(s) = -\pi \sin^3(\pi s) + 2\pi \sin(\pi s) \cos^2(\pi s) \).

Step 4: Evaluate \( g'(\frac{1}{2}) \)

Substitute \( s = \frac{1}{2} \) into \( g'(s) \). Evaluate: \( \sin(\pi \cdot \frac{1}{2}) = 1 \) and \( \cos(\pi \cdot \frac{1}{2}) = 0 \). This gives \( g'(\frac{1}{2}) = -\pi \cdot 1^3 + 2\pi \cdot 1 \cdot 0^2 = -\pi \).

Step 5: State the Result

The derivative evaluated at \( s = \frac{1}{2} \) is \( g'(\frac{1}{2}) = -\pi \).

Key Concepts

Understanding the Product Rule

When you have two functions multiplied together, like in this case with \( g(s) = \cos(\pi s) \sin^2(\pi s) \), the product rule is the tool you need to find the derivative. The product rule states: when differentiating two functions \( f(x) \) and \( h(x) \) that are multiplied, the derivative \( (fh)'(x) \) is given by \( f'(x) h(x) + f(x) h'(x) \).

In our original exercise, we divided the function into two parts, \( u = \cos(\pi s) \) and \( v = \sin^2(\pi s) \).

• The derivative of \( u \) is found to be \( u' = -\pi \sin(\pi s) \).
• Similarly, \( v' \) is found using the chain rule to be \( 2\pi \sin(\pi s) \cos(\pi s) \).

With these derivatives, apply the product rule by plugging \( u' \) and \( v' \) back into the formula: \( g'(s) = u'v + uv' \). This concept is crucial for handling products in derivatives.

Unlocking the Chain Rule

The chain rule is a fundamental tool for taking the derivative of composite functions, which are functions within other functions. It tells you how to differentiate a function that is embedded inside another function.

In our task, we applied the chain rule when differentiating \( v = \sin^2(\pi s) \). Here's how it works:

• You first differentiate the outer function — the square function into \( 2\sin(\pi s) \).
• Then, multiply it by the derivative of the inside function, the trigonometric function \( \sin(\pi s) \), which gives \( \pi \cos(\pi s) \).

Combine these results to arrive at \( v' = 2\pi \sin(\pi s) \cos(\pi s) \). This shows the inside-outside application that makes the chain rule unique. Mastering this rule will help you tackle more complex derivatives!

Insight into Trigonometric Functions

Trigonometric functions are a cornerstone in calculus. They have unique derivatives that are useful across many problems. In the expression \( \cos(\pi s) \sin^2(\pi s) \), these functions are at play, each influencing the outcome of the differentiation.

Knowing these derivatives is essential:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• For \( \cos(x) \), the derivative is \(-\sin(x)\).

In applying these derivatives, the constants like \( \pi \) are taken outside the differentiation. This means when differentiating \( \cos(\pi s) \), you multiply by \( \pi \), yielding \(-\pi \sin(\pi s)\). Similarly, handling \( \sin^2(\pi s) \) requires remembering its derivative is also guided by the chain rule. Knowing these trigonometric details gives you a deeper understanding of how derivatives operate with these functions.