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Chapter 4: Problem 10

1-44. Find the derivative of each function. $$ f(x)=\ln (5 x) $$

Short Answer

Expert verified
The derivative of \( f(x) = \ln(5x) \) is \( \frac{1}{x} \).

Step by step solution

01

Identify the Function and Its Components

We start with the function \( f(x) = \ln(5x) \), where \( \ln \) is the natural logarithm function and \( 5x \) is the argument inside the logarithm. To differentiate \( f(x) \), we'll use the chain rule since \( 5x \) is a phrase within the \( \ln \) function.
02

Recall the Derivative of the Logarithm Function

The derivative of \( \ln(u) \) with respect to \( x \), where \( u = g(x) \), is \( \frac{1}{u} \cdot \frac{du}{dx} \) according to the chain rule. Here, our \( u \) is \( 5x \), a linear function of \( x \).
03

Differentiate the Inner Function

We differentiate \( u = 5x \) with respect to \( x \). The derivative of \( 5x \) is a constant, simply \( \frac{du}{dx} = 5 \).
04

Apply the Chain Rule

Using the chain rule for logarithmic functions, substitute the derivatives we found, \( \frac{d}{dx} [\ln(5x)] = \frac{1}{5x} \cdot 5 \).
05

Simplify the Expression

Upon simplification, \( \frac{1}{5x} \cdot 5 = \frac{5}{5x} = \frac{1}{x} \). Thus, the derivative of the function \( f(x) = \ln(5x) \) is \( \frac{1}{x} \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental tool in calculus for finding the derivatives of composite functions. Think of it as a method for what happens when you have a function within another function. This rule helps you "peel back" the layers to differentiate each part.
  • First, identify the outer function and the inner function. In our case, for \( f(x) = \ln(5x) \), the outer function is \( \ln(u) \) and the inner function is \( u = 5x \).
  • The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
  • Here, we differentiate the outer function as if the inner part were just a variable, and then multiply by the derivative of the inner function.

When using the chain rule, always remember that the derivative of the outer function is evaluated at the inner function, and don't forget to multiply by the derivative of the inner function!
Natural Logarithm Function
The natural logarithm, denoted as \( \ln(x) \), is a special logarithm with base \( e \), where \( e \approx 2.71828 \). It is commonly used in calculus due to its unique properties, especially in differentiation and integration.
  • The derivative of \( \ln(u) \) is \( \frac{1}{u} \), where \( u \) is a function of \( x \).
  • This makes \( \ln(x) \) particularly powerful in calculations involving growth, decay, and compounding problems.

When differentiating a natural logarithm, always identify the argument inside the logarithm, represented as \( u \), and apply the derivative formula \( \frac{1}{u} \cdot \frac{du}{dx} \). This step simplifies complex expressions into manageable parts through differentiation.
Differentiation of Functions in Calculus
Differentiation in calculus is the process of finding the derivative of a function, which represents the rate of change of the function. Derivatives are foundational to understanding how functions behave.
  • The differentiation process involves rules like the chain rule, product rule, and quotient rule to simplify expressions and solve problems effectively.
  • When you differentiate, you essentially uncover the slope of the function at any given point or how the function's output changes with a slight change in input.
Understanding differentiation in calculus allows you to solve diverse problems in physics, engineering, and economics, illustrating how most functions change in the real world. Through practice, you become adept at applying the appropriate rules and simplifying expressions to find the derivatives of complex functions effectively.

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Most popular questions from this chapter

69-72. Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{x}+e^{-x} $$

\(45-50 .\) For each function, find the indicated expressions. $$ f(x)=x^{2} \ln x-x^{2}, \quad \text { find } \quad \text { a. } f^{\prime}(x) \quad \text { b. } f^{\prime}(e) $$

Solve the following exercises on a graphing calculator by graphing an appropriate exponential function (using \(x\) for ease of entry) together with a constant function and using INTERSECT to find where they meet. You will have to choose an appropriate window. A bank account grows at \(7 \%\) compounded continuously. How many years will it take to: a. double? b. increase by \(25 \% ?\)

\(63-68 .\) Find the differential of each function and evaluate it at the given values of \(x\) and \(d x\). $$ y=x e^{x} \text { at } x=1 \text { and } d x=0.1 $$

97-98. ATHLETICS: World's Record 100-Meter Run In 1987 Carl Lewis set a new world's record of 9.93 seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93 .\) Enter this function as \(y_{1},\) and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window [0,9.93] by [0,100] Trace along the velocity curve to verify that Lewis's maximum speed was about 11.27 meters per second. Find how quickly he reached a speed of 10 meters per second, which is \(95 \%\) of his maximum speed.
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