Problem 111 - Calculus Volume 1

Question

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=3 x\left(18 x^{4}+\frac{13}{x+1}\right)$$

Step-by-Step Solution

Answer

\( f'(x) = 270x^4 + \frac{39}{x+1} - \frac{39x}{(x+1)^2} \)

Step 1: Apply the Product Rule

The function given is of the form \( u(x) = 3x \) and \( v(x) = 18x^4 + \frac{13}{x+1} \). The derivative of \( f(x) = u(x)v(x) \) is given by the product rule: \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Step 2: Differentiate \( u(x) = 3x \)

The derivative of \( u(x) = 3x \) is \( u'(x) = 3 \).

Step 3: Differentiate \( v(x) = 18x^4 + \frac{13}{x+1} \)

First, find the derivative of \( 18x^4 \), which is \( 72x^3 \). Next, differentiate \( \frac{13}{x+1} \) using the quotient rule. Let \( v(x) = 13 \) and \( w(x) = x+1 \). Then, \( v'(x) = 0 \) and \( w'(x) = 1 \). The derivative is \(-\frac{13}{(x+1)^2} \). Thus, \( v'(x) = 72x^3 - \frac{13}{(x+1)^2} \).

Step 4: Substitute and Compute the Derivative

While using the product rule, substitute the derivatives into the formula: \[ f'(x) = 3\left(18x^4 + \frac{13}{x+1}\right) + 3x\left(72x^3 - \frac{13}{(x+1)^2}\right) \].

Step 5: Simplify the Expression

Simplify the expression: \[ f'(x) = 54x^4 + \frac{39}{x+1} + 216x^4 - \frac{39x}{(x+1)^2} \]. Combine like terms to finally get: \[ f'(x) = 270x^4 + \frac{39}{x+1} - \frac{39x}{(x+1)^2} \].

Key Concepts

Derivative

When you hear the term "derivative" in calculus, think about it as a way to find the rate at which something is changing. Derivatives provide information such as the slope of a curve at any point, which means how steep that curve grows or falls. Think of this like checking how fast a car is going at a specific moment.

In this exercise, the goal is to find the derivative of the function \(f(x) = 3x(18x^4 + \frac{13}{x+1})\). This function involves products of terms, making it suitable for the product rule. Each part of the function needs to be analyzed separately to find how they change before applying rules to combine them.

Once we've found how each part of the function changes separately, we can apply those insights to derive the overall rate of change for the whole expression. These calculations often involve applying rules of differentiation carefully and step-by-step so that no changes are missed.

Quotient Rule

The quotient rule is another vital tool in calculus used when you have divisions of two functions and you want to find out how the result of their division changes. Whenever you see a function given by one term divided by another, the quotient rule is the way to go. It compares how the top and bottom functions change to give an overall rate of change for their division.

For example, consider the term \(\frac{13}{x+1}\). Here, the numerator is a constant while the denominator is a simple linear function. Using the quotient rule,

• You consider the derivative of the numerator which is zero because it's a constant.
• Then, compute the derivative of the denominator, which is \(1\).
• Combine these results to get the full expression: \( -\frac{13}{(x+1)^2} \).

This tells us how the entire expression \(\frac{13}{x+1}\) changes with respect to \(x\), an integral piece when trying to find the derivative of more complex functions.

Simplifying Expressions

After calculating derivatives using product rule and quotient rule, there's usually still work to do: simplifying the obtained expression. Simplifying is about combining like terms and making the expression easier to understand or use. Think of it as cleaning up your workspace once you've done the complex work.

In this exercise, the derived expression comes out as: \[ f'(x) = 54x^4 + \frac{39}{x+1} + 216x^4 - \frac{39x}{(x+1)^2} \].

• Notice how some terms, like \(54x^4\) and \(216x^4\), can be combined because they involve the same power of \(x\).
• Combine these terms to simplify, resulting in \(270x^4\).

Additionally, reducing fractions or distributing common factors can also simplify the expression further. Simplifying doesn’t change the derivative. It just makes it look nicer, so it’s more practical for solving problems or interpretation.