Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the following. $$ y=(x+1)^{(x+1)}, \text { where } x>-1 \text { . } $$

Step-by-Step Solution

Answer

The derivative of the function \(y=(x+1)^{(x+1)}\) is \(\frac{dy}{dx}=(x+1)^x (x+2)\ln(x+1)+ (x+1)^x\)

Step 1: Convert the function form

First, convert the given function using logarithm. The natural log is taken for both sides of the equation. This gives a more manageable equation to differentiate. The initial function \(y=(x+1)^{(x+1)}\), is then written as \(\ln y=(x+1)\ln(x+1)\)

Step 2: Differentiate both sides

Performing the derivative of both sides with respect to \(x\), the derivative of the left side of the equation is obtained using the chain rule \(\frac{1}{y} \frac{dy}{dx}\). To accomplish the derivative of the right side of the equation, the product rule is used due to the multiplication between \((x+1)\) and \(\ln(x+1)\). This gives \(\frac{1}{y} \frac{dy}{dx}=\ln(x+1)+1\)

Step 3: Solve for dy/dx

To define \(\frac{dy}{dx}\), you simply multiply each side by \(y\). We express \(y\) as \((x+1)^{(x+1)}\) from the original function given. This yields \(\frac{dy}{dx}=((x+1)^{(x+1)}) (\ln(x+1)+1)\)

Step 4: Simplify dy/dx

Solving the equation, we get the derivative as \(\frac{dy}{dx}=(x+1)^x (x+2)\ln(x+1)+ (x+1)^x\)

Key Concepts

Chain Rule

The chain rule is a vital tool in differentiation, especially when dealing with composite functions. The chain rule helps you find the derivative of such functions by iteratively applying the simple rule that
the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Consider the expression \(y = f(g(x))\).The chain rule states that the derivative \(\frac{dy}{dx}\) is found by multiplying the derivative of the outer part \(f'(g(x))\) by the derivative of the inner part \(g'(x)\):\[\frac{dy}{dx} = f'(g(x)) \, \cdot \, g'(x)\]In the exercise, we applied the chain rule when differentiating the logarithmized form \(\ln y\).

• Given \(y\), we take \(\ln y\), transforming the differentiation task into \(\frac{1}{y} \frac{dy}{dx}\).
• The rule allows us to treat \(y\) as an implicit function with respect to \(x\).

This makes unraveling even complex functions more straightforward through traceable steps of differentiation.

Product Rule

When you encounter a function composed of two or more multiplicative parts that require differentiation, the product rule should come to mind.
This core rule helps simplify the process by breaking it down into understandable components, key for functions where two separate functions are multiplied together.The product rule states:

• If you have a function \(u(x)\times v(x)\), its derivative is
• \[ \frac{d}{dx}(u\cdot v) = u'\cdot v + u\cdot v' \]

In the context of our exercise:

• We saw this application in dealing with \((x+1)\ln(x+1)\). This involves differentiating "x+1" and "\ln(x+1)" separately.
• Using the product rule, the task becomes: \[\frac{d}{dx}((x+1)\ln(x+1)) = (1)\cdot \ln(x+1) + (x+1)\cdot \frac{1}{x+1}\]
• The solution, after simplifying, becomes the equation \(\ln(x+1) + 1\).

Thus, even with multiple parts in a function product, differentiation can be handled with clarity and precision.

Logarithmic Differentiation

Logarithmic differentiation is a strategy that simplifies differentiation, especially when facing functions with exponents or products.
It leverages the properties of logarithms to transform complex functions into more manageable forms.Here's how you can use it effectively:

• Convert to Logarithmic Form: Transform the equation by applying a natural log on both sides. For example, if you have \(y = (x+1)^{(x+1)}\), express it as \(\ln y = (x+1) \ln(x+1)\).
• Differentiate Log Form: Differentiating involves applying the chain rule on the left side and the product rule on the right side.
• Isolate \( \frac{dy}{dx} \): Multiply by \(y\) (back-substitute the original function) to clear any fractions, allowing you to solve for \( \frac{dy}{dx} \).

This method, as used in the exercise, helps transform and differentiate equations that look far more intimidating in their original form. It is particularly useful for functions where powers are functions of \(x\), resulting in a clean, actionable solution.