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Chapter 2: Problem 19

Find the derivative of each function. $$ f(r)=\pi r^{2} $$

Short Answer

Expert verified
The derivative of \( f(r) = \pi r^2 \) is \( f'(r) = 2\pi r \).

Step by step solution

01

Identify the Function Structure

The function given is \( f(r) = \pi r^2 \). Here, \( \pi \) is a constant and \( r^2 \) is a power of the variable \( r \). To find the derivative with respect to \( r \), we will apply the power rule of differentiation.
02

Apply the Power Rule

The power rule states that if \( f(r) = r^n \), then the derivative \( f'(r) = n \cdot r^{n-1} \). In our function, \( n = 2 \). Thus, the derivative of the \( r^2 \) term with respect to \( r \) is \( 2r \).
03

Incorporate the Constant Multiplier

When differentiating a function with a constant multiplier, the constant is preserved in the derivative. Our function is \( f(r) = \pi \cdot r^2 \). Since \( \pi \) is a constant, the derivative becomes \( \pi \cdot 2r \).
04

Simplify the Derivative

Simplify the expression derived from applying the rules. Multiply the constants together: \( f'(r) = 2 \pi r \). This is the derivative of the function \( f(r) \).

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Key Concepts

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

Power Rule
The Power Rule is a fundamental concept in calculus and makes finding derivatives much simpler. It states that if you have a function of the form \( f(x) = x^n \), where \( n \) is a constant, the derivative \( f'(x) \) is \( n \cdot x^{n-1} \). This rule allows you to "bring down" the exponent and reduce the power by one.
For example, consider our function \( f(r) = r^2 \). Using the Power Rule:
  • You "bring down" the \( 2 \) in front of \( r \).
  • Multiply the whole term by this number, resulting in \( 2 \cdot r^{2-1} \).
  • The exponent decreases by one, leaving \( 2r^1 \), or simply \( 2r \).
So, the Power Rule provides a quick and reliable method to find derivatives of polynomials and similar functions.
Constant Multiplier
In calculus, the Constant Multiplier Rule tells us how to handle constants when differentiating functions. If you have a constant multiplied by a function, the derivative is simply the constant times the derivative of the function.
When looking at the function \( f(r) = \pi r^2 \):
  • \( \pi \) is a constant, not a variable, meaning you leave it unchanged during differentiation.
  • First, differentiate \( r^2 \) using the Power Rule to get \( 2r \).
  • Then, multiply the constant by this derivative, giving \( \pi \cdot 2r \).
This approach ensures that constants are correctly incorporated into the derivative, avoiding any mistakes that might occur if they were ignored.
Function Derivatives
Function derivatives provide us with a way to understand how a function behaves with respect to its variable, indicating how the function changes at any given point. Differentiation is a powerful tool to analyze functions in mathematics.
Let's break it down using our function \( f(r) = \pi r^2 \):
  • The derivative \( f'(r) = 2\pi r \) gives the slope of the tangent line to the curve at any point \( r \).
  • This means if you know \( r \), you can determine in what manner and how fast \( f(r) \) is changing.
  • Such information is crucial in physics, engineering, economics, and anywhere rates of change are essential.
Through the process of differentiation, we acquire a deeper understanding of the dynamic nature of functions and their derivatives.

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Most popular questions from this chapter

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\left[\left(x^{3}+1\right)^{2}-x\right]^{4} $$

Find \(f^{\prime}(x)\) by using the definition of the derivative. [Hint: See Example 4.] $$ \text { } \begin{aligned} f(x)=x^{5} & \\ &\left[\text { Hint: Use } \quad(x+h)^{5}=\right.\\\ &\left.x^{5}+5 x^{4} h+10 x^{3} h^{2}+10 x^{2} h^{3}+5 x h^{4}+h^{5}\right] \end{aligned} $$

A company's cost function is \(C(x)=\sqrt{4 x^{2}+900}\) dollars, where \(x\) is the number of units. Find the marginal cost function and evaluate it at \(x=20\).

The Chain Rule (page 155 ) states that the derivative of \(f(g(x))\) is \(f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Use Carath茅odory's definition of the derivative to prove the Chain Rule by giving reasons for the following steps. a. Since \(g\) is differentiable at \(x\), there is a function \(G\) that is continuous at 0 and such that \(g(x+h)-g(x)=G(h) \cdot h\), and \(G(0)=g^{\prime}(x)\). b. Since \(f\) is differentiable at \(g(x)\), there is a function \(F\) that is continuous at 0 and such that \(f(g(x)+h)-f(g(x))=F(h) \cdot h\), and \(F(0)=f^{\prime}(g(x))\) c. For the function \(f(g(x))\) we have \(f(g(x+h))-f(g(x))\) \(\quad=f(g(x)+g(x+h)-g(x))-f(g(x))\) \(\quad=f(g(x)+(g(x+h)-g(x)))-f(g(x))\) \(=F(g(x+h)-g(x)) \cdot(g(x+h)-g(x))\) \(=F(g(x+h)-g(x)) \cdot G(h) \cdot h\) Therefore, the derivative of \(f(g(x))\) is \(\begin{aligned} F(g(x+0)-g(x)) \cdot G(0) &=F(0) \cdot G(0) \\ &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \end{aligned}\) as was to be proved.

Describe the difference between a secant line and a tangent line for the graph of a function. What formula would you use to find the slope of the secant? What formula for the tangent?
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