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Chapter 4: Problem 49

Use logarithmic differentiation to differentiate the following functions. $$f(x)=x^{x}$$

Short Answer

Expert verified
The derivative of $$x^x$$ is $$x^x (\text{ln}(x) + 1)$$.

Step by step solution

01

Identify the function and apply natural logarithm

Given the function is $$f(x) = x^x$$. To differentiate this function using logarithmic differentiation, start by taking the natural logarithm of both sides: $$\text{ln} \big(f(x)\big) = \text{ln} \big( x^x \big)$$.
02

Simplify using logarithm properties

Use the property of logarithms that $$\text{ln} (a^b) = b \text{ln}(a)$$ to simplify the equation: $$\text{ln} \big(f(x) \big) = x \text{ln}(x)$$.
03

Differentiate both sides implicitly

Differentiate both sides of the equation with respect to $$x$$. On the left side, use the chain rule: \[ \frac{d}{dx} \big( \text{ln}( f(x)) \big) = \frac{1}{f(x)} \frac{d}{dx} f(x) \ \text{On the right side, use product rule:} \ \frac{d}{dx} \big( x \text{ln}(x) \big) = \text{ln}(x) + 1 \ \text{Putting it all together:} \ \frac{1}{f(x)} \frac{d}{dx} f(x) = \text{ln}(x) + 1 \ \text{Since} \ f(x) = x^x \ \text{we get:} \ \frac{1}{x^x} \frac{d}{dx} x^x = \text{ln}(x) + 1 \ \text{implies:} \ \frac{d}{dx} x^x = x^x (\text{ln}(x) + 1 ) \].

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Key Concepts

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

Natural Logarithm
To understand logarithmic differentiation, you first need to grasp the concept of a natural logarithm. The natural logarithm, denoted as \text{ln}(x)\, is the logarithm to the base \(e\), where \(e\) is an irrational number approximately equal to 2.718. Natural logarithms have some interesting properties, one of which simplifies the differentiation process of complex functions like \(f(x) = x^x\).
Product Rule
When differentiation involves a product of two functions, you need to use the product rule. The product rule states that for two functions \(u(x)\) and \(v(x)\), the derivative of their product is \frac{d}{dx} [u(x) \times v(x)] = u'(x) \times v(x) + u(x) \times v'(x)\. In the given example, we encounter this in the step where we differentiate \(x \times \text{ln}(x)\). When applying the product rule to \(x \text{ln}(x)\), we treat \(x\) as \(u(x)\) and \text{ln}(x)\ as \(v(x)\). Therefore: \frac{d}{dx} [x \times \text{ln}(x)] = 1 \times \text{ln}(x) + x \times \frac{1}{x} = \text{ln}(x) + 1\
Chain Rule
The chain rule is a method for finding the derivative of a composition of functions. If you have a function \(y = g(f(x))\), the chain rule states that the derivative of \(y\) with respect to \(x\) is the derivative of \(g\) with respect to \(f\) times the derivative of \(f\) with respect to \(x\): \frac{d}{dx} g(f(x)) = g'(f(x)) \times f'(x)\. In the logarithmic differentiation process of \(f(x) = x^x\), when differentiating \(\text{ln}(f(x))\) with respect to \(x\), we use the chain rule: \frac{d}{dx} \big(\text{ln}(f(x))\big) = \frac{1}{f(x)} \times \frac{d}{dx} f(x)\

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Most popular questions from this chapter

Differentiate the following functions. $$y=8 e^{x}\left(1+2 e^{x}\right)^{2}$$

Calculate values of \(\frac{10^{x}-1}{x}\) for small values of \(x\), and use them to estimate \(\left.\frac{d}{d x}\left(10^{x}\right)\right|_{x=0} .\) What is the formula for \(\frac{d}{d x}\left(10^{x}\right) ?\)

Find the equation of the tangent line to the graph of \(y=e^{x}\) at \(x=0 .\) Then, graph the function and the tangent line together to confirm that your answer is correct.

Graph the function \(y=\ln \left(e^{x}\right),\) and use trace to convince yourself that it is the same as the function \(y=x\). What do you observe about the graph of \(y=e^{\ln x} ?\)

Find the values of \(x\) at which the function has a possible relative maximum or minimum point. (Recall that \(e^{x}\) is positive for all \(x .\) ) Use the second derivative to determine the nature of the function at these points. $$f(x)=(1+x) e^{-3 x}$$
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