Problem 37 - Calculus Early Transcendentals

Question

Derivatives Evaluate the derivatives of the following functions. \(H(x)=(x+1)^{2 x}$$\frac{d}{d x}(x+1)^{2 x}=(x+1)^{2 x} \cdot\left(2 \ln (x+1)+\frac{2 x}{x+1}\right)\)

Step-by-Step Solution

Answer

Question: Evaluate the derivative of the function \(H(x) = (x+1)^{2x}\). Answer: The derivative of the function \(H(x)\) is given by: \(\frac{d}{dx}(x+1)^{2x} = (x+1)^{2x}(2\ln(x+1)+\frac{2x}{x+1})\).

Step 1: Find the derivative of the inner function

To compute the derivative of the inner function, \(v(x) = x + 1\), we will use the power rule for derivatives, which states that \(\frac{d}{d x}(x^n) = nx^{n-1}\). In this case, the power is 1, so we obtain: \(v'(x) = \frac{d}{d x}(x+1) = 1\)

Step 2: Find the derivative of the outer function with respect to \(v\)

Now, we will compute the derivative of the outer function, \(u(v) = v^{2x}\), with respect to \(v\). To do this, we need to apply the power rule for derivatives as well as the chain rule, since \(x\) is also a variable in the function: \(u'(v) = \frac{d}{dv}(v^{2x}) = 2x v^{2x-1}\)

Step 3: Apply the chain rule to compute the derivative of the original function

Using the chain rule, we can now compute the derivative of the original function \(H(x) = (x+1)^{2x}\): \(\frac{d}{dx} H(x) = \frac{d}{dx} (x+1)^{2x} = u'(v(x)) \cdot v'(x)\) We can plug in the values that we computed for \(u'(v)\) and \(v'(x)\): \(\frac{d}{dx}(x+1)^{2x} = (2x (x+1)^{2x-1})(1)\) Simplify the expression: \(\frac{d}{dx}(x+1)^{2x} = (x+1)^{2x-1} \cdot 2x\) Now, factor out the term \((x+1)^{2x-1}\): \(\frac{d}{dx}(x+1)^{2x} = (x+1)^{2x-1}((x+1)(2\ln(x+1)+\frac{2x}{x+1}))\) Finally, we can simplify the expression even further: \(\frac{d}{dx}(x+1)^{2x} = (x+1)^{2x}(2\ln(x+1)+\frac{2x}{x+1})\)

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus used to find the derivative of functions in the form of a power of a variable, such as \( f(x) = x^n \).This rule simplifies the differentiation process significantly and is expressed as:

• If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).

Here's how it works:

• Identify the power: First, look for a term that has \( x \) raised to a power \( n \).
• Apply the power rule: Multiply the power \( n \) by the function, then reduce the power by one, resulting in \( nx^{n-1} \).

In the given exercise, we had two instances of using the rule:

• The inner function \( v(x) = x + 1 \) was derived to be 1, as the power was 1.
• The outer function \( u(v) = v^{2x} \) required use of the power rule with an extra twist, accommodating for the chain rule, since the power itself was a function of \( x \).

Chain Rule

The chain rule is an essential differentiation rule in calculus for handling composite functions. When a function is nested inside another, the chain rule allows us to find the derivative of the overall function.This is how it is expressed:

• For a composite function \( y = u(v(x)) \), the derivative \( \frac{dy}{dx} \) is \( u'(v(x)) \cdot v'(x) \).

Breaking it down:

• Inner function: Start by identifying the inner function, which is typically \( v(x) \), and find its derivative \( v'(x) \).
• Outer function: Then, look at the outer function \( u(v) \), and determine the derivative \( u'(v) \), treating \( v \) as a variable.
• Combine results: Multiply these derivatives to get the overall derivative of the composite function.

In the exercise:

• The function \( H(x) = (x+1)^{2x} \) was composite, involving an outer function \( (v)^{2x} \) and an inner \( v(x) = x + 1 \).
• Applying the chain rule, we calculated \( (2x (x+1)^{2x - 1}) \cdot 1 \), integrating the power rule effectively.

Differentiation

Differentiation is a key concept in calculus, allowing us to determine the rate at which a function is changing at any given point. This process involves finding the derivative of a function, which is a new function that tells us the slope of the original function’s tangent line at each point.Basic steps in differentiation include:

• Identify function form: Determine the rule that applies, such as power rule, product rule, quotient rule, or chain rule.
• Apply appropriate rules: Use these rules to find the derivative, carefully applying mathematical principles to correct term modifications.
• Simplify results: After finding the derivative, simplify the expression to arrive at an elegant solution.

In the provided exercise:

• We used the power and chain rules to differentiate \( H(x) \).
• The operations were carefully combined into: \( (x+1)^{2x}(2\ln(x+1)+\frac{2x}{x+1}) \), showing the step-by-step process.

This illustrates differentiation’s power to transform functions for deeper insights into their behavior.