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Chapter 4: Problem 76

$$ \text { Differentiate. } $$ $$ f(x)=\frac{1}{5} x^{5}\left(\ln x-\frac{1}{5}\right) $$

Short Answer

Expert verified
The derivative is \(f' (x) = x^4 \ln x\).

Step by step solution

01

- Identify the Product Rule

The function is a product of two functions: \(u = \frac{1}{5} x^5\) and \(v = \ln x - \frac{1}{5}\). We need to apply the product rule: \[\frac{d}{dx}[u \, v] = u' \, v + u \, v'.\]
02

- Differentiate the First Function

Find the derivative of \(u = \frac{1}{5} x^5\). \[u' = \frac{d}{dx} \left( \frac{1}{5} x^5 \right) = x^4.\]
03

- Differentiate the Second Function

Find the derivative of \(v = \ln x - \frac{1}{5}\). \[v' = \frac{d}{dx} \left( \ln x - \frac{1}{5} \right) = \frac{1}{x}.\]
04

- Apply the Product Rule

Use the product rule to find \(f' (x)\): \[f' (x) = u' v + u \, v' = x^4 \left( \ln x - \frac{1}{5} \right) + \frac{1}{5} x^5 \left( \frac{1}{x} \right).\]
05

- Simplify the Expression

Simplify the expression obtained: \[f' (x) = x^4 \ln x - \frac{1}{5} x^4 + \frac{1}{5} x^4 = x^4 \ln x.\]

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Key Concepts

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

Differentiation
Differentiation helps us find how a function changes as its input changes. Think of it as a way to measure the rate at which something is changing. In the given exercise, we want to find the derivative, which tells us the slope or rate of change of the function at any point. Differentiating a function helps us understand how it behaves and can tell us if it's increasing, decreasing, or changing at a particular rate. For example, in the given exercise, we start by splitting the function into parts and then find the derivative of each part.
Product Rule
When differentiating a product of two functions, we need the product rule. The product rule formula is \( \frac{d}{dx}[u \, v] = u' \, v + u \, v'\).

In the solution, we have two functions: \( u = \frac{1}{5} x^5 \) and \( v = \ln x - \frac{1}{5} \). This formula helps us find the derivative as follows:

* Differentiate \(u\), so \(u' = x^4\).
* Differentiate \(v\), so \(v' = \frac{1}{x}\).
* Apply the product rule: \( f'(x) = u'v + uv'\).
Therefore, \( f'(x) = x^4 ( \ln x - \frac{1}{5}) + \frac{1}{5} x^5 (\frac{1}{x}) \). This gives us the rate of change of the function.
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a special logarithm with base \( e \), where \( e \) is approximately 2.71828. It's a key concept in calculus because its properties make differentiation simpler. When differentiating \( \ln x \), the result is \( \frac{1}{x} \).

In our exercise, the function \( v \) contains \( \ln x - \frac{1}{5} \). Differentiating this using the properties of logarithms gives \( v' = \frac{1}{x} \). This makes our calculations straightforward and aligns perfectly with the product rule.

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Most popular questions from this chapter

The annual consumption of pork per person has declined from \(52 \mathrm{lb}\) in 1980 to \(48 \mathrm{lb}\) in 2000 . Assume consumption is decreasing according to the exponential-decay model. \({ }^{25}\) a) Find the value of \(k\) and write the equation. b) Estimate the consumption of pork in \(2010 .\) c) In what year (theoretically) will the consumption of pork be \(10 \mathrm{lb}\) per person?

In an art class, students were tested at the end of the course on a final exam. Then they were retested with an equivalent test at subsequent time intervals. Their scores after time \(t\), in months, are given in the following table. $$ \begin{array}{|c|c|} \hline \text { Time, } t \text { (in months) } & \text { Score, } y \\ \hline 1 & 84.9 \% \\ 2 & 84.6 \% \\ 3 & 84.4 \% \\ 4 & 84.2 \% \\ 5 & 84.1 \% \\ 6 & 83.9 \% \\ \hline \end{array} $$ a) Use the REGRESSION feature on a grapher to fit a logarithmic function \(y=a+b \ln x\) to the data. b) Use the function to predict test scores after \(8 \mathrm{mo} ; 10 \mathrm{mo} ; 24 \mathrm{mo} ; 36 \mathrm{mo}\) c) After how long will the test scores fall below \(82 \% ?\) d) Find the rate of change of the scores and interpret its meaning.

The coroner arrives at the scene of a murder at \(11 \mathrm{P.M}\). She takes the temperature of the body and [inds it to be \(85.9^{\circ}\). She waits \(1 \mathrm{hr}\), takes the temperature again, and finds it to be \(83.4^{\circ}\). She notes that the room temperature is \(60^{\circ} .\) When was the murder committed?

Differentiate. $$ f(x)=10^{x} $$

Bornstein and Bornstein found in a study that the average walking speed \(v\) of a person living in a city of population \(p\), in thousands, is given by $$ v(p)=0.37 \ln p+0.05 $$ where \(v\) is in feet per second. \({ }^{12}\) a) The population of Seattle is \(531,000 .\) What is the average walking speed of a person living in Seattle? b) The population of New York is \(7,900,000\). What is the average walking speed of a person living in New York? c) Find \(v^{\prime}(p)\). d) Interpret \(v^{\prime}(p)\) found in part (c).
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