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Chapter 3: Problem 38

In Exercises \(37-42,\) find \(f^{\prime}(x)\) and state the domain of \(f^{\prime}\) $$f(x)=\ln (2 x+2)$$

Short Answer

Expert verified
The derivative of the function is \(f^{\prime}(x) = \frac{1}{x+1}\) and its domain is \(x \in (-\infty, -1) \cup (-1, +\infty)\).

Step by step solution

01

Apply the Chain Rule to Find the Derivative

The derivative of the function \(f(x)=\ln (2 x+2)\) can be calculated using the chain rule, which states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. The derivative of a logarithmic function \(\ln u\), where \(u\) is a function of \(x\), can be written as \(\frac{1}{u}\). Therefore, the derivative of the function is \(f^{\prime}(x) = \frac{1}{2x+2}\times\frac{d(2x+2)}{dx}.\)
02

Calculate the Derivative of the Inside Function

The inside function is \(2x+2\). Its derivative, calculated using the power rule, is \(\frac{d(2x+2)}{dx}=2\). Substituting this in the previous equation gives \(f^{\prime}(x) = \frac{1}{2x+2}\times 2\).
03

Simplify the Derivative

After simplifying, the derivative of the function becomes \(f^{\prime}(x) = \frac{2}{2x+2} = \frac{1}{x+1}.\)
04

Find the Domain of the Derivative

The derivative is defined when the denominator is not equal to zero. Thus we set \(x+1 \neq 0\), giving \(x \neq -1\). Therefore, the domain of the derivative is \(-\infty<x<-1\) and \(-1<x<\infty\), which is the same as \(x \in (-\infty, -1) \cup (-1, +\infty)\).

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Key Concepts

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

Understanding the Chain Rule
The chain rule is a fundamental concept in calculus, particularly when dealing with composite functions. Composite functions can be tricky because they involve a function nested inside another. To handle these, the chain rule tells us how to take derivatives effectively.
  • The chain rule states that to find the derivative of a composite function \(g(f(x))\), you need to multiply the derivative of the outer function \(g\) evaluated at the inner function \(f\), by the derivative of the inner function \(f(x)\).
  • In our exercise, the function is \(f(x) = \ln(2x + 2)\), where \(\ln(x)\) is the outer function and \(2x + 2\) is the inner function.
  • The derivative of \(\ln(u)\) is \(\frac{1}{u}\). So, applying the chain rule involves multiplying \(\frac{1}{2x + 2}\) by the derivative of the inner function \(2x + 2\), which is 2.
Therefore, the derivative \(f'(x)\) becomes \(\frac{2}{2x + 2}\). Simplifying this expression gives \(\frac{1}{x + 1}\). The chain rule simplifies the process of derivatives for composite functions and is incredibly useful in calculus.
Delving into Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, and they play a key role when handling growth and decay problems in various fields.
  • One of the most common logarithmic functions you will encounter in calculus is the natural logarithm, denoted as \(\ln(x)\), which is the inverse of the exponential function \(e^x\).
  • The derivative of the natural logarithm \(\ln(x)\) is quite simple: it's \(\frac{1}{x}\). This property is integral when using the chain rule, as seen in the context of our exercise: \(f(x) = \ln(2x + 2)\).
The function \(f(x)\) becomes differentiable by taking advantage of the simple rule that the derivative of \(\ln(u)\) is \(\frac{1}{u}\). This means we can easily take the derivative by identifying the inner function and applying the chain rule. Understanding how logarithmic and exponential functions relate makes it easier to manipulate their properties, such as derivatives.
Exploring the Domain of a Function
The domain of a function is essentially the set of all possible inputs (or \(x\)-values) for which the function is defined. Determining the domain is crucial, especially when evaluating functions involving logarithms or fractions, because certain inputs can make the function undefined.
  • For our derivative \(f'(x) = \frac{1}{x+1}\), the function is undefined when the denominator equals zero.
  • This means the function doesn鈥檛 exist where \(x + 1 = 0\), simplifying to \(x = -1\).
Hence, the domain of \(f'(x)\) is all real numbers except for \(x = -1\). Expressed in interval notation, that鈥檚 \((-\infty, -1) \cup (-1, +\infty)\). Knowing the domain helps avoid undefined expressions and ensures continuity where possible.

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Most popular questions from this chapter

In Exercises 74 and \(75,\) use the curve defined by the parametric equations \(x=t-\cos t, y=-1+\sin t\) Multiple Choice Which of the following is an equation of the tangent line to the curve at \(t=0 ?\) (A) \(y=x\) (B) \(y=-x\) (C) \(y=x+2\) (D) \(y=x-2 \quad(\) E) \(y=-x-2\)

True or False The acceleration of a particle is the second derivative of the position function. Justify your answer.

Multiple Choice Which of the following is \(\frac{d}{d x} \sec ^{-1}\left(x^{2}\right) ?\) (A) \(\frac{2}{x \sqrt{x^{4}-1}} \quad\) (B) \(\frac{2}{x \sqrt{x^{2}-1}} \quad\) (C) \(\frac{2}{x \sqrt{1-x^{4}}}\) (D) \(\frac{2}{x \sqrt{1-x^{2}}} \quad\) (E) \(\frac{2 x}{\sqrt{1-x^{4}}}\)

In Exercises \(37-42,\) find \(f^{\prime}(x)\) and state the domain of \(f^{\prime}\) $$f(x)=\ln (x+2)$$

True or False The slope of \(x y^{2}+x=1\) at \((1 / 2,1)\) is \(2 .\) Justify your answer.
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