Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 23 through 29, differentiate. In Problems 23 through 25, assume \(f\) is differentiable. Your answers may be in terms of \(f\) and \(f^{\prime} .\) $$ f(x)=\ln \left(e^{(x+5)^{2}}\right) $$

Step-by-Step Solution

Answer

The derivative of the given function is \(f'(x) = 2x+10\).

Step 1: Simplify the function

Before differentiating, there is a chance to simplify the function. As the function \(f(x)\) is the natural logarithm of the exponential function (which are inverse functions), they cancel each other, simplifying the function to \(f(x) = (x+5)^2\).

Step 2: Apply chain rule

Now differentiate using the chain rule \(d/dx [u(v)] = u'(v) \cdot v'\) , where \(u(v) = (v)^2\) and \(v=x+5\). Hence, the derivative \(f'(x)\) of \(f(x)\) is given by \(f'(x) = 2 v \cdot v'\).

Step 3: Differentiate the inner function

The derivative (with respect to \(x\)) of the inner function \(v = x+5\) is \(v' = 1\). Thus, substitute \(v'\) into the equation from step 2.

Step 4: Substitute and simplify

Substitute \(v'\) and \(v = x+5\) into the equation from step 2. Therefore, \(f'(x) = 2 (x+5) \cdot 1\). This simplifies to \(f'(x)=2x+10\).

Key Concepts

Inverse Functions

Inverse functions are a fascinating aspect of mathematics that essentially "undo" each other. Think of them as pairs where one function does something, and the other reverses it. For instance, the exponential function and the natural logarithm are inverse functions.
When you have a function like \(f(x) = \ln(e^{(x+5)^2})\), the natural logarithm \(\ln\) and the exponential function \(e\) cancel each other out. This makes the function simpler and is a key step before differentiation.
To better understand, remember these points:

• Inverse functions swap the role of input and output.
• If \(f(g(x)) = x\) and \(g(f(x)) = x\), then \(f\) and \(g\) are inverses.
• In solving problems, using inverses can greatly simplify expressions, as seen in this example where it reduced to \((x+5)^2\).

Understanding inverse functions can save time and effort, especially when dealing with complex expressions.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It helps us find the derivative of a function that is made up of other functions. It is especially useful when dealing with functions like \((x+5)^2\).
The basic idea is to take the derivative of the outer function and multiply it by the derivative of the inner function:

• Take \(u(v)\), the outer function as \((v)^2\).
• Set \(v\), the inner function, as \(x+5\).
• Apply the chain rule: \(f'(x) = 2v \cdot v'\), where \(v'\) is the derivative of the inner function.

For example, if \(v = x+5\), then \(v' = 1\). This makes calculating the derivative straightforward. Multiplying the derivatives together helps in finding the rate of change of complicated functions efficiently.

Simplifying Expressions

Simplifying expressions is a crucial skill in mathematics that makes it easier to handle and solve problems. Let's break down why it's beneficial, using our example:
The original function was \(f(x) = \ln(e^{(x+5)^2})\), which might look daunting at first. However, recognizing that \(\ln(e^x) = x\) simplifies it to \((x+5)^2\), which is much easier to work with.
Simplifying expressions involves:

• Identifying potential simplifications like cancellations or like terms.
• Breaking down complex expressions to more manageable forms.
• Using known identities and inverse relationships, such as logarithms and exponentials.

This simplification helps not just in differentiation but also in other mathematical operations, making computation less error-prone and more intuitive.