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Chapter 3: Problem 61

Use logarithmic differentiation to evaluate $f^{\prime}(x)$$ $$f(x)=\frac{(x+1)^{10}}{(2 x-4)^{8}}$$

Short Answer

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Question: Find the derivative of the function \(f(x)=\frac{(x+1)^{10}}{(2x-4)^8}\) using logarithmic differentiation. Answer: \(f^{\prime}(x)= \frac{(x+1)^9}{(2x-4)^8}(10-\frac{8(x+1)}{2x-4})\)

Step by step solution

01

Take the natural logarithm of both sides.

To begin, we'll take the natural logarithm of both sides of the equation \(f(x)=\frac{(x+1)^{10}}{(2x-4)^8}\): $$\ln(f(x))=\ln\left(\frac{(x+1)^{10}}{(2x-4)^8}\right)$$
02

Simplify the logarithm using properties.

Next, use the properties of logarithms to simplify the right side of the equation. Recall that \(\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)\) and \(\ln(a^n)=n\ln(a)\). Applying these properties, we get: $$\ln(f(x))=10\ln(x+1)-8\ln(2x-4)$$
03

Differentiate both sides.

Now, differentiate both sides of the equation with respect to \(x\). On the left side, use the chain rule: \(\frac{d}{dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)}\). On the right side, apply the chain rule and product rule: $$\frac{f^{\prime}(x)}{f(x)}=10\frac{d}{dx}\ln(x+1)-8\frac{d}{dx}\ln(2x-4)$$ $$\frac{f^{\prime}(x)}{f(x)}=\frac{10}{x+1}-\frac{16}{2x-4}$$
04

Solve for \(f^{\prime}(x)\).

Finally, multiply both sides of the equation by \(f(x)\) to solve for \(f^{\prime}(x)\): $$f^{\prime}(x)=\frac{(x+1)^{10}}{(2x-4)^8}\left[\frac{10}{x+1}-\frac{16}{2x-4}\right]$$
05

Simplify the expression for \(f^{\prime}(x)\).

Simplify the expression for \(f^{\prime}(x)\) by reducing common factors in the numerators and denominators: $$f^{\prime}(x)=\frac{10(x+1)^9}{(2x-4)^8} - \frac{8(2x-4)(x+1)^{10}}{(2x-4)^9} = \frac{10(x+1)^9}{(2x-4)^8} - \frac{8(x+1)^{10}}{(2x-4)^8}$$ Now, you can factor out the common term of \(\frac{(x+1)^9}{(2x-4)^8}\) to simplify further: $$f^{\prime}(x)= \frac{(x+1)^9}{(2x-4)^8}(10-\frac{8(x+1)}{2x-4})$$ This is the evaluated derivative \(f^{\prime}(x)\) using logarithmic differentiation.

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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 vital tool in calculus used to differentiate composite functions. It helps us find the derivative of functions nested within one another. When faced with a function of a function, like \(g(h(x))\), the chain rule allows us to differentiate it sequentially.

In this context, the chain rule states:
  • If a function \(y = f(u)\) and \(u = g(x)\), then the derivative \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
In our exercise, the chain rule was used to handle the natural logarithms like \( \ln(f(x)) \). We differentiated \(\ln(a^n)\) by pulling down the exponent \(n\), and applying the chain rule to find the rate of change with respect to \(x\). This makes it simpler to work with expressions involving nested and complex functions.
Product Rule
The product rule comes into play when differentiating expressions where two functions are multiplied together. The rule helps us understand how changes in each function affect the overall expression.

According to the product rule:
  • If \(y = u(x) \cdot v(x)\), then \(\frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)\).
In the logarithmic differentiation exercise, using the product rule might seem less evident, as we're dealing directly with division and constants. Nonetheless, the product rule's principles are implicit when simplifying the derivative expressions after applying properties of logarithms. It highlights the contributions of each separate component to the overall derivative.
Properties of Logarithms
Logarithms have unique properties that simplify mathematical expressions and play a crucial role in logarithmic differentiation.

These properties include:
  • \(\ln(a \cdot b) = \ln(a) + \ln(b)\)
  • \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
  • \(\ln(a^n) = n \cdot \ln(a)\)
In the exercise, these properties helped to break down a complex rational function into a set of operations that are simpler to differentiate. Instead of directly differentiating \(f(x)\) as a quotient, these logarithm properties allowed us to differentiate each component separately, greatly simplifying the problem.
Simplifying Expressions
Simplifying expressions is the process of making them easier to understand and solve. In calculus, simplifying an expression can involve factoring, combining like terms, or breaking down complex fractions.

In this exercise, after applying the chain rule and properties of logarithms, it was crucial to simplify the derivative \(f'(x)\) for practical utility.
  • Factoring out common terms: This helped to condense terms to a manageable format.
  • Reducing fractions: Allowed us to identify and cancel out factors.
The simplification process is essential not just for derivations but for verifying the correctness of the result. It ensures the derivative is expressed in its simplest form and lays a clearer path for subsequent calculations or evaluations.

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