Problem 30 - Calculus Early Transcendentals

Question

Find the indicated derivative. $$ \frac{d C}{d r}, \text { where } C=2 \pi r $$

Step-by-Step Solution

Answer

The derivative is \( 2\pi \).

Step 1: Identify the Function

First, let's recognize that the function for which we need to find the derivative is given by \( C = 2\pi r \). This implies that \( C \) is expressed as a function of \( r \) with constants multiplying \( r \).

Step 2: Apply Derivative Rules

The expression \( 2\pi r \) is a linear function of \( r \). The derivative of a constant times a function is simply the constant times the derivative of the function itself. Here, \( 2\pi \) is a constant.

Step 3: Compute the Derivative

The derivative of \( r \) with respect to \( r \) is \( 1 \). Hence, apply this to our function: \( \frac{d}{d r}(2 \pi r) = 2\pi \cdot 1 = 2\pi \).

Step 4: Combine and Simplify

Thus, the derivative of the function \( C \) with respect to \( r \) is \( 2\pi \). This completes the differentiation process whereby the constant \( 2\pi \) remains as the derivative since we differentiated with respect to \( r \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. A composite function is essentially a function within another function. To apply the chain rule, you identify the "outer" function and the "inner" function. It's crucial for solving complex differentiation problems. For example, if you have a function of the form \( f(g(x)) \), where \( f \) is the outer function and \( g \) is the inner function, the chain rule states that the derivative \( f'(g(x)) \) is the product of the derivative of the outer function evaluated at the inner function, and the derivative of the inner function.

• Start by differentiating the outer function with respect to the inner function.
• Then, differentiate the inner function with respect to the original variable.
• Finally, multiply these two derivatives together.

This method allows you to handle derivatives involving nested functions effectively. It's like peeling an onion, where you work from the outside to the inside.

Derivative of Linear Function

Linear functions are the simplest forms of functions to differentiate. A linear function can generally be written in the form \( f(x) = ax + b \). Here, \( a \) and \( b \) are constants, and \( a \) represents the slope of the function—the rate at which the function increases or decreases. Differentiating a linear function is straightforward because each component's derivative is constant.

• The derivative of a constant (i.e., \( b \) in the function) is zero.
• The derivative of \( ax \), with respect to \( x \), is simply \( a \). This is because \( ax \) is a linear term, and the differentiation reduces it to its coefficient.

In the exercise, to find the derivative of \( C = 2\pi r \), we identify \( 2\pi \) as a constant, and the derivative of \( r \) with respect to \( r \) is \( 1 \). Thus, multiplying, you obtain \( 2\pi \), which is a straightforward outcome of applying these basic rules.

Calculus Basics

Calculus is the mathematical study of change, and it consists primarily of differential calculus and integral calculus. Differential calculus involves finding the rate at which something changes, which is captured through differentiation. This is crucial for understanding how functions behave and are valuable for many applications, from physics to economics.

• Derivatives: A derivative represents the rate of change of a function with respect to a variable. It's like measuring how steeply a function ascends or descends at a specific point.
• Simplification: Often, calculus encourages simplification of expressions to find derivatives, like recognizing linear terms or constants.
• Applications: Practically, derivatives help in finding slopes of curves, optimizing real-world problems, and understanding dynamic systems.

The exercise showcases these basics through a real-world scenario—finding the derivative of a linear function, which is a fundamental process that lays the groundwork for more advanced calculus concepts. Being fluent in these basics makes tackling complex problems much more manageable.