Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(f^{\prime}(x) .\) Do these problems without using the Quotient Rule. $$ f(x)=\frac{4}{\sqrt{e^{x}+1}} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \frac{4}{\sqrt{e^{x}+1}}\) is \(f^{\prime}(x) = -2e^{x}(e^{x} +1)^{-3/2}\).

Step 1: Simplify the function

Rearrange the function as follows: \(f(x) = 4(e^{x} +1)^{-1/2}\). The step may not seem necessary but it simplifies the process of applying the chain rule and helps to avoid using the quotient rule.

Step 2: Apply the Chain Rule

The Chain Rule states that the derivative of a composition of functions is the derivative of the outer function multiplied by the derivative of the inner function. Let's treat \(4\) as a constant and the part \((e^{x} +1)^{-1/2}\) as \(u^{-1/2}\), where \(u = e^{x} + 1\). Then differentiate the function as normally would for a power function, bringing the exponent in front and reducing the exponent by one: \(f^{\prime}(x) = 4 * -1/2 (e^{x} +1)^{-3/2} = -2 (e^{x} +1)^{-3/2}.\)

Step 3: Differentiate the inner function

Now differentiate the inner function \(u = e^{x} + 1\). The derivative for \(e^{x}\) is \(e^{x}\) and for \(1\) is \(0\). So, compared to the initial \((e^{x} + 1)^{-1/2}\), the derivative becomes \((e^{x} + 1)^{-3/2} * e^{x}\).

Step 4: Combine the results

Now combine the results from the last two steps, multiplicate the result from the outer function with the derivative of the inner function: \(f^{\prime}(x) = -2 (e^{x} +1)^{-3/2} * e^{x} = -2e^{x}(e^{x} +1)^{-3/2}.\)

Key Concepts

Understanding the Chain Rule

The Chain Rule is a powerful tool in calculus, helping us find the derivative of composite functions. A composite function is like a "function within a function" situation. For instance, if you have a function like \( f(g(x)) \), you are dealing with two layers of functions: \( f \) on the outside and \( g \) nested inside.

To differentiate such functions, the Chain Rule comes to our rescue. It tells us to follow these steps:

• First, differentiate the outer function while keeping the inner function intact.
• Then, multiply this result by the derivative of the inner function.

This might seem a tad complex at first, but breaking it down can make it digestible. For example, when you have \( (e^x + 1)^{-1/2} \), you differentiate it as a power function. The exponent \( -1/2 \) comes down, and you decrease it by one to get \( -3/2 \). You then multiply the power expression by the derivative of \( e^x + 1 \). It is important to be thoughtful with each step and practice with various functions to master this rule.

Differentiating Exponential Functions

Exponential functions always have the form \( a^x \) and, whether due to popularity or their application, are a common sight in calculus. When you are working with these, remember that for the natural exponential function \( e^x \), its derivative is, interestingly, itself. In simpler terms, the slope of \( e^x \) at any point is equal to the value of \( e^x \) at that point.

Consider when you apply the chain rule to a function like \((e^x + 1)^{-1/2}\). Differentiating \( e^x \) yields \( e^x \), while the constant \( 1 \) vanishes since a constant's derivative is zero. Combining this result with the Chain Rule steps results in a smooth calculation process. Understanding these rules removes the complexity and eases the calculation of derivatives for such functions.

Importance of Function Simplification

Before jumping into applying complicated calculus rules, foresee if simplifying the function will make the process smoother. Function simplification can often dodge cumbersome rules and provide clarity in what needs to be done. For instance, transforming \( \frac{4}{\sqrt{e^x+1}} \) into \( 4(e^x +1)^{-1/2} \) allowed for a more straightforward application of the Chain Rule.

Simplification can involve:

• Reworking roots and powers when relating them to fractions.
• Combining like terms, which might reduce or eliminate unnecessary complexities.

This ensures you work with a manageable expression, reducing chances for errors and making the differentiation steps clearer. Developing a knack for it comes with practice and will significantly aid in distributing focus for accurate differentiation.