Problem 37 - Calculus Early Transcendentals

Question

Evaluate the derivatives of the following functions. $$H(x)=(x+1)^{2 x}$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function $$H(x)=(x+1)^{2 x}$$. Answer: The derivative of the function $$H(x)=(x+1)^{2 x}$$ is $$\frac{dH}{dx}=x(1+x)^{x-1}$$.

Step 1: Apply the Chain Rule

Recall that the Chain Rule states that if we have a composite function $$H(x)=g(f(x))$$, then the derivative of H with respect to x is the product of the derivative of g with respect to f(x) and the derivative of f with respect to x, or: $$\frac{dH}{dx}=\frac{dg}{df(x)}\cdot\frac{df(x)}{dx}$$ In our case, we can rewrite the function as: $$H(x)=g(f(x))=((1+x)^{2})^x$$

Step 2: Identify the Inner and Outer Functions

Now identify the inner and outer functions: $$f(x)=1+x$$ $$g(u)=u^x, \:\mbox{where}\: u=f(x)=1+x$$

Step 3: Find the Derivative of the Inner Function

The derivative of f(x) with respect to x is the derivative of $$1+x$$ which is simply: $$\frac{df(x)}{dx}=1$$

Step 4: Find the Derivative of the Outer Function

Use the power rule to find the derivative of g(u) with respect to u, remembering that x is a constant in this case: $$\frac{dg(u)}{du}=xu^{x-1}$$ Now, substitute back f(x) for u: $$\frac{dg(u)}{du}=x(1+x)^{x-1}$$

Step 5: Apply the Chain Rule

Using the Chain Rule as defined in Step 1, we multiply the derivative of the inner function and the outer function to find the derivative of H(x): $$\frac{dH}{dx}=\frac{dg}{df(x)}\cdot\frac{df(x)}{dx}=x(1+x)^{x-1}\cdot 1 =x(1+x)^{x-1}$$ Thus, the derivative of the function $$H(x)=(x+1)^{2 x}$$ is: $$\frac{dH}{dx}=x(1+x)^{x-1}$$

Key Concepts

Composite Functions

When dealing with calculus, especially when taking derivatives, the concept of **Composite Functions** is key. Think of composite functions as functions within functions. Imagine peeling an onion; you have layers, and that’s similar to the layers of functions in composites.
A composite function is expressed as \( y = g(f(x)) \), where \( f(x) \) is the inner function and \( g(u) \) is the outer function. For example, if we have \( H(x) = ((1+x)^{2})^x \), it can be rewritten as \( H(x) = g(f(x)) \) where:

• \( f(x) = 1 + x \)
• \( g(u) = u^x \) with \( u = f(x) \)

This splitting makes it easier to take derivatives, unlocking the power of the Chain Rule. Remember, composite functions allow you to handle intricate equations by dissecting them into manageable parts.

Power Rule

The **Power Rule** is one of the simplest but most powerful tools in calculus, especially when differentiating functions. The rule is straightforward: if you have a function \( y = x^n \), the derivative, with respect to \( x \), would be \( y' = nx^{n-1} \).
In the context of composite functions, the Power Rule is essential. If you have an outer function like \( g(u) = u^x \), where \( u \) is another function of \( x \), then you'll employ the Power Rule to find \( \frac{dg}{du} \).
This involves treating \( x \) as a constant while differentiating. So, if \( g(u) = (1+x)^2 \), the derivative here involves using the rule such that \( g(u) = xu^{x-1} \). It’s this application of the Power Rule that helps simplify even the complex composite expressions.

Derivative Calculation

In calculus, computing derivatives is fundamental. The **Derivative Calculation** in our context involves using both the Chain Rule and the Power Rule.
The Chain Rule is essential in linking together the derivative of composite functions. For example, if you have \( H(x) = ((1+x)^{2})^x \), you first identify:

• Inner function \( f(x) = 1+x \)
• Outer function \( g(u) = u^x \)

Then you take the derivative of \( f(x) \), which is 1, and the derivative of \( g(u) \) with respect to \( u \) using the Power Rule, giving you \( xu^{x-1} \). Finally, substitute back \( f(x) \) for \( u \) and multiply by the derivative of \( f(x) \) to get \( x(1+x)^{x-1} \).
Putting this all together, the derivative of \( H(x) \) provides insights into the rate of change of the function, vital for understanding the behavior of functions in calculus.