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Chapter 8: Problem 5

\(\frac{\mathrm{d}}{\mathrm{d} x} \ln \left(\frac{x+1}{2 x}\right)\) is: (a) \(\frac{1}{2}\) (b) \(\frac{1}{x+1}-\frac{1}{2 x}\) (c) \(\frac{2 x}{x+1}\) (d) \(\frac{1}{x+1}+\frac{1}{x}\) (e) \(\frac{1}{x+1}-\frac{1}{x}\).

Short Answer

Expert verified
Option (e): \( \frac{1}{x+1} - \frac{1}{x} \).

Step by step solution

01

Rewrite the Logarithmic Expression

Use the properties of logarithms to rewrite the expression inside the natural logarithm. The property to use is: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). Therefore, \( \ln\left(\frac{x+1}{2x}\right) = \ln(x+1) - \ln(2x) \).
02

Differentiate the Logarithms Individually

Differentiate each term using the chain rule. The derivative of \( \ln(x+1) \) is \( \frac{1}{x+1} \), and the derivative of \( \ln(2x) \) is \( \frac{1}{2x} \cdot \frac{d}{dx}(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).
03

Combine the Results

Subtract the derivatives obtained in the previous step. This gives: \( \frac{1}{x+1} - \frac{1}{x} \).
04

Identify the Correct Answer

Compare the result \( \frac{1}{x+1} - \frac{1}{x} \) with the given choices. It matches option (e).

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Key Concepts

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

Natural Logarithm Properties
The natural logarithm, often written as \text{ln}(x)\, is a special logarithm with the base of Euler's number \(e\approx 2.718\). One of the key properties used in solving logarithmic differentiation problems is the logarithmic property that relates the logarithm of a quotient to the difference of logarithms.
  • \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
In the given exercise, this property is used to simplify \( \ln\left(\frac{x+1}{2x}\right)\) into two separate terms:
\[ \ln\left(\frac{x+1}{2x}\right) = \ln(x+1) - \ln(2x) \]
This makes differentiating the expression significantly easier in subsequent steps.
Chain Rule
The chain rule is an essential principle in calculus, used to differentiate composite functions. When you have a function of a function, the chain rule helps you find the derivative.
Consider a function \(y = f(g(x))\). According to the chain rule:
  • \(\frac{dy}{dx} = f'(g(x)) \, g'(x)\)
In the context of our exercise, the chain rule is applied in differentiating \(\ln(x+1)\) and \(\ln(2x)\). First, we identify the inner functions (inside the logarithms), and then we find the derivatives of these inner functions:
  • The derivative of \(\ln(x+1)\) is \(\frac{1}{x+1}\).
  • The derivative of \(\ln(2x)\) involves first differentiating \(2x\) which is \(2\), then using the chain rule to find \(\frac{1}{2x} \, \cdot \, 2 = \frac{1}{x}\).
Differentiation of Logarithmic Functions
Differentiating logarithmic functions like \(\ln(x)\) involves specific properties and rules. For a general logarithmic function \(\ln(u)\), where \(u\) is a differentiable function of \(x\), the derivative is given by:
  • \(\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}\)
In the exercise, the terms \(\ln(x+1)\) and \(\ln(2x)\) were differentiated separately:
  • For \(\ln(x+1)\), the derivative is \(\frac{1}{x+1}\).
  • For \(\ln(2x)\), we apply the chain rule to get \( \frac{1}{2x} \, \cdot \, 2 = \frac{1}{x} \).
This differentiation forms the basis for obtaining the final result by combining these derivatives:
\ \[ \frac{1}{x+1} - \frac{1}{x} \]This matches the given choice (e).

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Most popular questions from this chapter

\(\frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^{x^{2}+1}\right)\) is: (a) \(2 x\) (b) \(2 x \mathrm{e}^{x^{2}+1}\) (c) \(2 x \mathrm{e}^{2 x}\) (d) \(\left(x^{2}+1\right) \mathrm{e}^{x^{2}}\) (e) \(\ln \left(x^{2}+1\right) \mathrm{e}^{x^{2}+1}\).

Show that the graph of \(y=\frac{a x+b}{c x+d}\) has, in general, no turning points and that $$ 2\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)\left(\frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}\right)=3\left(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right)^{2} $$

If \(\mathrm{e}^{x}=7.3\), the value of \(x\), to 2 s.f. is: (a) \(2.0\) (b) \(1.0\) (c) \(-2.0\) (d) 0 (e) \(0.2\).

\(\frac{\mathrm{d}}{\mathrm{d} x}\left[\ln \frac{x}{1+x}\right]=\frac{\mathrm{d}}{\mathrm{d} x}[\ln x]-\frac{\mathrm{d}}{\mathrm{d} x}[\ln (1+x)]\).

If \(x^{2}+y^{2}=4\) then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is: (a) \(2 x+2 y\) (b) \(4-x^{2}\) (c) \(-\frac{x}{y}\) (d) \(\frac{y}{x}\) (e) \(\frac{4-x}{y}\).
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