Problem 55 - Calculus Early Transcendentals

Question

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(1+\frac{4}{x}\right)^{x}$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function $$y = \left(1+\frac{4}{x}\right)^{x}$$ with respect to $$x$$. Answer: The derivative of the function is $$\frac{d}{dx}\left(1+\frac{4}{x}\right)^{x} = \left(1+\frac{4}{x}\right)^{x}\left(\ln{\left(1+\frac{4}{x}\right)} - \frac{4}{x+\frac{4}{x}}}\right)$$.

Step 1: Rewrite the function with a natural logarithm

Let $$y = \left(1+\frac{4}{x}\right)^{x}$$. Take the natural logarithm of both sides: $$\ln{y} = \ln{\left(1+\frac{4}{x}\right)^{x}}$$.

Step 2: Simplify the equation

Use the logarithmic rule $$\ln a^b = b \ln a$$ to simplify the equation: $$\ln{y} = x \ln{\left(1+\frac{4}{x}\right)}$$.

Step 3: Implicit differentiation

Differentiate both sides of the simplified equation with respect to $$x$$: $$\frac{d}{dx}(\ln{y}) = \frac{d}{dx}\left[x \ln{\left(1+\frac{4}{x}\right)}\right]$$.

Step 4: Apply the product rule and chain rule

Using the product rule on the right-hand side and the chain rule for the logarithm: $$\frac{1}{y}\frac{dy}{dx} = \ln{\left(1+\frac{4}{x}\right)} + x\left(\frac{-4}{x^2}\right)\frac{1}{1+\frac{4}{x}}$$.

Step 5: Solve for the derivative

Multiply both sides by $$y$$ to solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \left(1+\frac{4}{x}\right)^{x}\left(\ln{\left(1+\frac{4}{x}\right)} - \frac{4}{x+\frac{4}{x}}}$$. So, the derivative of the given function is: $$\frac{d}{dx}\left(1+\frac{4}{x}\right)^{x} = \left(1+\frac{4}{x}\right)^{x}\left(\ln{\left(1+\frac{4}{x}\right)} - \frac{4}{x+\frac{4}{x}}}\right)$$

Key Concepts

Derivatives

A derivative represents how a function changes as its input changes. Simply put, it measures the rate at which something is changing. For instance, if you're driving, the derivative of your position with respect to time is your speed.

When working with derivatives, you're often given a function and asked to find how it behaves as it changes. This involves applying different rules like the product rule or chain rule, which we will explore more below. In practical terms, finding a derivative can help you understand slopes, tangents, and even accelerate calculations in various fields of science and engineering.

Product Rule

The product rule is a vital tool in calculus when finding derivatives of products of two functions. When you have a function that is the product of two other functions, \[ u(x) = f(x) imes g(x) \] there's a specific way to find its derivative.

According to the product rule, the derivative is given by:\[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]

Here's how it works in real terms:

• Differentiate the first function while keeping the second constant.
• Then, differentiate the second function while keeping the first constant.
• Add these two results together for the final derivative.

Chain Rule

The chain rule is essential for dealing with composite functions, which are functions composed of other functions. Suppose you have a function, \[ h(x) = f(g(x)) \].

The chain rule says you can find its derivative by:\[ h'(x) = f'(g(x)) imes g'(x) \]

This might sound complex, but think of it step-by-step:

• First, differentiate the outer function while leaving the inner function unchanged.
• Second, multiply this by the derivative of the inner function.
• This "chain" of derivatives helps unravel complicated expressions.

In the original step-by-step solution, you see the chain rule in action when handling the composition within the logarithmic expression.

Logarithmic Differentiation

Logarithmic differentiation is a clever trick to deal with more complex functions involving exponents. When you apply logarithms to both sides of an equation, you simplify the computation of derivatives. Applying the natural log (ln) helps break down powers into manageable pieces using the property:\[ \ln(a^b) = b\ln(a) \]

After applying the logarithm, you proceed to differentiate implicitly. This involves treating the function inside the log normally while remembering to multiply the derivative by the original function at the end.

• Start by taking the natural logarithm of both sides.
• Simplify the expression using logarithmic identities.
• Differentiate implicitly like normal but remember to multiply by the original function after.

This method can simplify the process of finding derivatives of complicated functions, making it easier to handle expressions with variable exponents.