Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-32, 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 at \( s = \frac{1}{2} \) is \( -\pi \).

Step 1: Identify the Function

The problem provides the function \( g(s) = \cos(\pi s) \sin^2(\pi s) \). We need to find the derivative of this function, specifically at \( s = \frac{1}{2} \).

Step 2: Apply the Product Rule

The function \( g(s) = \cos(\pi s) \sin^2(\pi s) \) is a product of two functions: \( u(s) = \cos(\pi s) \) and \( v(s) = \sin^2(\pi s) \). Using the product rule, \( (uv)' = u'v + uv' \), we find the derivatives of both parts.

Step 3: Differentiate \( u(s) = \cos(\pi s) \)

Differentiate \( u(s) \) with respect to \( s \):\[ u'(s) = \frac{d}{ds}[\cos(\pi s)] = -\pi \sin(\pi s) \] using the chain rule.

Step 4: Differentiate \( v(s) = \sin^2(\pi s) \)

Differentiate \( v(s) = \sin^2(\pi s) \) using the chain rule:\[ v'(s) = 2 \sin(\pi s) \cdot \cos(\pi s) \cdot \pi = 2\pi \sin(\pi s)\cos(\pi s) \] using the double angle identity, it becomes \( \pi \sin(2\pi s) \).

Step 5: Apply the Product Rule

Using the product rule, the derivative \( g'(s) \) is:\[ g'(s) = u'(s)v(s) + u(s)v'(s) = (-\pi \sin(\pi s)) \sin^2(\pi s) + \cos(\pi s)(\pi \sin(2\pi s)) \] simplify to approximate values if needed to calculate at a specific point.

Step 6: Evaluate at \( s = \frac{1}{2} \)

Substitute \( s = \frac{1}{2} \) into the derivative:\[ g'\left(\frac{1}{2}\right) = [ -(\pi) \sin\left(\frac{\pi}{2}\right) \cdot \sin^2\left(\frac{\pi}{2}\right)] + \cos\left(\frac{\pi}{2}\right) \cdot \pi \sin(\pi) = -(\pi) \cdot 1 \cdot 1 + 0 \cdot 0 = -\pi \]

Step 7: State the Final Answer

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

Key Concepts

Product Rule

In calculus, the product rule is an essential tool for differentiating functions that are multiplied together. If you have a product of two functions, say \( u(s) \) and \( v(s) \), the derivative of their product is given by the formula: \( (uv)' = u'v + uv' \). This means that you differentiate the first function while keeping the second one unchanged, and then add the result to the first function unchanged while differentiating the second one.For instance, consider our function \( g(s) = \cos(\pi s) \sin^2(\pi s) \). Here, \( u(s) = \cos(\pi s) \) and \( v(s) = \sin^2(\pi s) \). Using the product rule, we have

• \( u'(s) = -\pi \sin(\pi s) \)
• \( v'(s) = \pi \sin(2\pi s) \)

Therefore, substituting these into the product rule gives us \( g'(s) = u'(s)v(s) + u(s)v'(s) \), which can then be simplified to find the derivative of the entire expression.

Chain Rule

The chain rule is another vital theorem in calculus used to differentiate composite functions. When you have a function inside of another function, you can think of the chain rule as a way to "unroll" these functions. Given a function \( h(s) = f(g(s)) \), the derivative using the chain rule can be expressed as: \( h'(s) = f'(g(s)) \cdot g'(s) \). In our problem, we used the chain rule while differentiating both \( \cos(\pi s) \) and \( \sin^2(\pi s) \). For example:

• When differentiating \( \cos(\pi s) \), we treat \( \pi s \) as an inner function. So we obtain \( u'(s) = -\pi \sin(\pi s) \).
• Similarly, when differentiating \( \sin^2(\pi s) \), we apply the chain rule, recognizing \( \pi s \) and \( \sin(\pi s) \) as inner functions: \( v'(s) = 2\pi \sin(\pi s) \cos(\pi s) = \pi \sin(2\pi s) \).

By implementing the chain rule, you ensure that you account for every layer of functions nested within each other.

Trigonometric Functions

Trigonometric functions such as sine and cosine are periodic functions essential in calculus, especially when dealing with oscillatory behaviors. These functions have specific derivative rules that are helpful to memorize:

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

In the original function \( g(s) = \cos(\pi s) \sin^2(\pi s) \), we see these trigonometric functions playing a significant role. When applying both the product rule and chain rule, it's essential to use these derivative rules to break down the problem correctly.By understanding these trigonometric identities and their derivatives, you can manage their appearances in more complex calculus problems. As seen with terms like \( \sin(2\pi s) \), understanding trigonometric identities simplifies the differentiation of functions that involve squared terms or multiple angles.