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Logarithmic differentiation 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: \(f^{\prime}(x)=\frac{10(x+1)^9}{(2x-4)^8}-\frac{16(x+1)^{10}}{(2x-4)^9}\)

Step by step solution

01

Take the natural logarithm of both sides

Take the natural logarithm of both sides of the function definition. Remember that for ln, the natural logarithm function, $$\ln(a^n)=n\ln (a)$$ Applying that property, we get: $$\ln (f(x))=\ln\left(\frac{(x+1)^{10}}{(2x-4)^8}\right)$$
02

Use the properties of logarithms to expand

We will expand this expression using the properties of logarithms. Specifically, we will apply the quotient rule and power rule: 1. Quotient rule: \(\ln(\frac{a}{b})=\ln(a)-\ln(b)\) 2. Power rule: \(\ln(a^b)=b\ln(a)\) Applying those rules, we get: $$\ln (f(x))=10\ln(x+1)-8\ln(2x-4)$$
03

Differentiate implicitly with respect to x

Now, we will differentiate both sides of the equation with respect to x. Keep in mind that, $$\frac{d}{dx} \ln{(u)}=\frac{u'}{u}$$ Applying that differentiation rule, we get: $$\frac{f'(x)}{f(x)}=10\frac{1}{(x+1)}-8\frac{2}{(2x-4)}$$
04

Multiply both sides by f(x) and substitute original function

Multiply both sides by \(f(x)\): $$f'(x)=10\frac{1}{(x+1)}f(x)-8\frac{2}{(2x-4)}f(x)$$ Substitute \(f(x)=\frac{(x+1)^{10}}{(2 x-4)^{8}}\) to solve for \(f^{\prime}(x)\): $$f^{\prime}(x)=10\frac{1}{(x+1)}\frac{(x+1)^{10}}{(2x-4)^8}-8\frac{2}{(2x-4)}\frac{(x+1)^{10}}{(2x-4)^8}$$
05

Simplify the expression

Simplify the expression to obtain the final solution: $$f^{\prime}(x)=\frac{10(x+1)^9}{(2x-4)^8}-\frac{16(x+1)^{10}}{(2x-4)^9}$$ So, the derivative of the function \(f(x)\) using logarithmic differentiation is: $$f^{\prime}(x)=\frac{10(x+1)^9}{(2x-4)^8}-\frac{16(x+1)^{10}}{(2x-4)^9}$$

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a powerful technique used when dealing with functions in which one variable is not isolated. Instead of requiring the function to be explicit, this method allows us to differentiate implicitly by treating one variable as a function of another.
For example, in logarithmic differentiation, we often apply implicit differentiation to differentiate both sides of an equation.

Here's how it works:
  • Differentiation is performed with respect to one variable (usually denoted as \( x \)).
  • Every time you differentiate a term involving a function of \( x \), it is treated as a function, and implicit differentiation is applied.
  • This technique is especially useful when differentiating complicated composite functions.
In our example, implicit differentiation is applied after using properties of logarithms to simplify the logarithmic expression before differentiating it with respect to \( x \). By doing so, we can find the derivative without explicitly solving the function for \( x \).
Derivative of Logarithmic Functions
The derivative of logarithmic functions allows us to handle expressions involving logarithms effectively. Logarithmic differentiation is especially helpful when functions are products, quotients, or powers, as seen in complicated functions.
The key rule for differentiation of a natural log (\( \ln \)) function is:
  • \( \frac{d}{dx} \ln(u) = \frac{u'}{u} \)
where \( u \) is a differentiable function of \( x \) and \( u' \) is its derivative.

This rule essentially transforms the process of differentiating complex products or quotients by converting them into sums or differences that are easier to handle.
In the exercise, the application of this rule is pivotal in differentiating the expanded logarithmic expressions derived from the original function.
By taking the derivative of these logs, we convert multiplicative and divisive relationships in the function to additive and subtractive, streamlining the differentiation process.
Properties of Logarithms
The properties of logarithms, like the Power Rule and the Quotient Rule, are essential tools in simplifying complex logarithmic expressions. These properties allow us to rewrite logarithmic functions in more manageable forms, which is valuable during differentiation.

Here's a quick recap of the important properties of logarithms:
  • Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
  • Power Rule: \( \ln(a^b) = b\ln(a) \)
These rules enable us to break down a fraction or a power in the argument of the log function into sums or differences.
In the provided exercise, applying these properties allowed us to express the natural log of the given function in terms of the natural logs of its components.
This transformation simplifies the differentiation process by removing the complexity introduced by exponential or fractional terms. Understanding and applying these properties is crucial for making logarithmic differentiation less burdensome and more straightforward.

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Most popular questions from this chapter

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$e^{x y}=2 y$$

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x+y^{3}-x y=1\) (Hint: Rewrite as \(y^{3}-1=x y-x\) and then factor both sides.)

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$(x+y)^{2 / 3}=y ;(4,4)$$

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. Find \(d y / d t\), the velocity of the mass. Assume \(k\) and \(m\) are constant. b. How would the velocity be affected if the experiment were repeated with four times the mass on the end of the spring? c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness ( \(k\) is increased by a factor of 4 )? d. Assume y has units of meters, \(t\) has units of seconds, \(m\) has units of kg, and \(k\) has units of \(\mathrm{kg} / \mathrm{s}^{2} .\) Show that the units of the velocity in part (a) are consistent.

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