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Chapter 2: Problem 32

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

Short Answer

Expert verified
The derivative \( g'\left( \frac{1}{2} \right) \) is \(-\pi\).

Step by step solution

01

Identify the Function to Differentiate

The function we want to take the derivative of is given by \( g(s) = \cos(\pi s) \sin^2(\pi s) \). We are asked to find \( g'\left( \frac{1}{2} \right) \).
02

Use the Product Rule

To differentiate \( g(s) \), apply the product rule for the function \( u(s) = \cos(\pi s) \) and \( v(s) = \sin^2(\pi s) \). The product rule is \( (uv)' = u'v + uv' \).
03

Differentiate \( u(s) = \cos(\pi s) \)

The derivative \( u'(s) \) is \( -\pi \sin(\pi s) \) because the derivative of \( \cos(x) \) is \( -\sin(x) \), and we use the chain rule since the argument is \( \pi s \).
04

Differentiate \( v(s) = \sin^2(\pi s) \) using the Chain Rule

Rewrite \( v(s) \) as \( (\sin(\pi s))^2 \). Using the chain rule, let \( w(t) = t^2 \) and \( t = \sin(\pi s) \), so \( v'(s) = 2 \cdot \sin(\pi s) \cdot \pi \cos(\pi s) \).
05

Substitute the Derivatives Into the Product Rule Formula

Substitute \( u'(s) \) and \( v'(s) \) into the product rule formula: \[g'(s) = (-\pi \sin(\pi s))\sin^2(\pi s) + \cos(\pi s)(2\sin(\pi s) \cdot \pi \cos(\pi s)).\]
06

Simplify the Expression

Simplify the expression: \[g'(s) = -\pi \sin(\pi s) \cdot \sin(\pi s) \cdot \sin(\pi s) + 2\pi \cos^2(\pi s) \cdot \sin(\pi s).\]Rewrite the first term as \( -\pi \sin^3(\pi s) \) and the second term as \( 2\pi \sin(\pi s) \cos^2(\pi s) \).
07

Evaluate the Derivative at \( s = \frac{1}{2} \)

Substitute \( s = \frac{1}{2} \): \[g'\left( \frac{1}{2} \right) = -\pi \sin^3\left(\frac{\pi}{2}\right) + 2\pi \sin\left( \frac{\pi}{2} \right) \cos^2\left( \frac{\pi}{2} \right).\]Since \( \sin\left( \frac{\pi}{2} \right) = 1 \) and \( \cos\left( \frac{\pi}{2} \right) = 0 \), the expression simplifies to \( -\pi \).
08

Simplify to Get the Final Result

After substituting and simplifying, the derivative at \( s = \frac{1}{2} \) is:\[g'\left( \frac{1}{2} \right) = -\pi.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
A key technique in calculus for differentiating two multiplied functions is the Product Rule. It's extraordinarily helpful when confronted with a function that is the product of two separate functions. The rule is stated as: if you have two functions, say, \( u(x) \) and \( v(x) \), then the derivative of their product \( uv \) is calculated as \((uv)' = u'v + uv'\). This means that to differentiate \( uv \), you need to find the derivative of \( u \) and multiply it by \( v \), then add it to the product of \( u \) and the derivative of \( v \). This rule is a cornerstone of differentiation techniques and makes handling complex products straightforward.
Chain Rule
The Chain Rule is an essential differentiation technique used when dealing with composite functions鈥攆unctions within functions. Consider a function \( f(g(x)) \), where \( g \) is wrapped in \( f \). The Chain Rule states that the derivative of \( f(g(x)) \) is \( f'(g(x)) \, g'(x) \). This means you differentiate the outer function \( f \) with respect to \( g \) and multiply by the derivative of the inner function \( g \) with respect to \( x \). This chaining allows us to peel back layers of complexity in functions, seamlessly managing nested relationships.
Trigonometric Functions
Trigonometric functions like sine and cosine are central to a vast array of problems in calculus. Differentiating these functions requires recognition of some key patterns:
  • The derivative of \( \sin(x) \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) is \( -\sin(x) \).
When these functions involve more complex arguments, like \( \pi s \) instead of just \( s \), you employ the chain rule. This allows for differentiation of functions like \( \cos(\pi s) \) by acknowledging the factor of \( \pi \) through the derivative of its inner expression, resulting in \( -\pi \sin(\pi s) \). These patterns are invaluable tools.
Differentiation Techniques
Differentiation techniques are the methods used to find the rate at which a function changes. In calculus, these techniques are crucial for understanding how to navigate more complex expressions:
  • The Product Rule is applied when differentiating products of functions.
  • The Chain Rule is employed for nested functions.
  • Special rules exist for trigonometric functions.
Through these techniques, one can break down complicated expressions into manageable parts, find derivatives efficiently, and solve real-world problems where rates of change are crucial. Understanding these methods empowers you to handle a wide variety of calculus challenges.

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Most popular questions from this chapter

Consider \(y=x^{3}-1\). (a) Sketch its graph as carefully as you can. (b) Draw the tangent line at (2,7) . (c) Estimate the slope of this tangent line. (d) Calculate the slope of the secant line through (2,7) and \(\left(2.01,(2.01)^{3}-1.0\right)\) (e) Find by the limit process (see Example 1) the slope of the tangent line at (2,7)

An object travels along a line so that its position \(s\) is \(s=t^{2}+1\) meters after \(t\) seconds. (a) What is its average velocity on the interval \(2 \leq t \leq 3 ?\) (b) What is its average velocity on the interval \(2 \leq t \leq 2.003 ?\) (c) What is its average velocity on the interval \(2 \leq t \leq 2+h ?\) (d) Find its instantaneous velocity at \(t=2\).

Find \(D_{x} y .\) \(y=x^{2} \cos x\)

Suppose that an object moves along a coordinate line so that its directed distance from the origin after \(t\) seconds is \(\sqrt{2 t+1}\) feet. (a) Find its instantaneous velocity at \(t=\alpha, \alpha>0\). (b) When will it reach a velocity of \(\frac{1}{2}\) foot per second? (see Example \(5 .)\)

Use \(f^{\prime}(x)=\lim _{t \rightarrow x}[f(t)-f(x)] /[t-x]\) to find \(f^{\prime}(x)\). \(f(x)=x^{2}-3 x\)
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