Question

In Exercises, find the derivative of the function. $$ f(x)=\frac{2}{\left(e^{x}+e^{-x}\right)^{3}} $$

Short answer

The derivative of the function \(f(x) = 2x \ln x\) is \(f'(x) = 2 + 2\ln x\).

Step-by-step solution

Identify the Functions

The function \(f(x) = 2x \ln x\) is a product of two simpler functions. Identify these two functions as \(u(x) = 2x\) and \(v(x) = \ln x\).

Find the Derivatives of the Functions

Next, compute the derivatives of \(u(x)\) and \(v(x)\). The derivative of \(u(x) = 2x\) is \(u'(x) = 2\), and the derivative of \(v(x) = \ln x\) is \(v'(x) = 1/x\).

Apply the Product Rule

Apply the product rule of differentiation, which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. This gives \(f'(x) = u(x)v'(x) + v(x)u'(x) = 2x * 1/x + \ln x * 2\).

Simplify the Result

Simplify the resulting expression to get the final answer. This gives \(f'(x) = 2 + 2\ln x\).

Key concepts

Product Rule

When solving calculus problems that involve the product of two functions, the product rule is an essential technique. The product rule helps us find the derivative of two or more multiplied functions. This is crucial because the derivative of a product is not merely the product of the derivatives, as some might initially think.

The product rule formula is:

• If you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x) \, v(x) \) is expressed as:
• \( (u \, v)' = u' \, v + u \, v' \)

This formula shows us that, to find the derivative of a product, you take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function. This concept is particularly vital in a wide range of mathematical applications where functions are operating jointly, such as in our example of \( f(x)=2x \ln x \).

By applying the product rule to \( f(x)=2x \ln x \), first identify \( u(x) = 2x \) and \( v(x) = \ln x \). The derivative of \( f(x) \), using the product rule, will result from the combination of performing operations with \( u(x) \) and \( v(x) \), and their respective derivatives.

Logarithmic Function

The logarithmic function, especially the natural logarithm \( \ln x \), plays a pivotal role in calculus and mathematical analysis. Logarithms are the inverse operations to exponentiation, helping us manipulate equations involving exponential growth phenomena.

There's a specific reason we use base \( e \) (about 2.71828), calling it the natural logarithm. This base has properties that simplify the derivative process. The derivative of the natural logarithm \( \ln x \) with respect to \( x \) is particularly straightforward:

• \( \frac{d}{dx}\ln x = \frac{1}{x} \)

This simplicity is why the natural logarithm is favored in differentiation problems like the one we discussed. Working with the \( \ln x \) function in calculus-based problems allows for straightforward manipulation and integration, which can be highly advantageous.

In the example exercise, recognizing \( v(x) = \ln x \) allowed us to quickly determine its derivative as \( v'(x) = 1/x \), simplifying our process of applying the product rule and computing the final derivative of the function.

Differentiation

Differentiation is a core concept in calculus, which involves finding the rate at which a function is changing at any given point. It's essentially the process of finding a derivative.

Derivatives can tell us a lot about the behavior of functions:

• They tell us about the slope of the tangent line to the curve of the function at any point.
• They provide information about increasing or decreasing intervals.
• They are foundational in solving complex calculus problems such as optimization and motion analysis.

In our sample problem, differentiation is used to extract further information from the initial function \( f(x) = 2x \ln x \). By differentiating, we determine \( f'(x) = 2 + 2\ln x \), which provides insights into the slope of the function at various points.

Mastering differentiation, particularly how to apply rules like the product rule in various contexts, equips one with the tools necessary to tackle advanced mathematical challenges. The function \( f(x) = 2x \ln x \), a combination of polynomial and logarithmic components, illustrates how diverse operations converge in calculus, highlighting the synergy between different functions.