Problem 54 - Brief Applied Calculus

Question

Find the derivative of each function. $$ \left(x^{5}+1\right) \frac{x^{3}+2}{x+1} $$

Step-by-Step Solution

Answer

Use product and quotient rules to find the derivative: \( (uv)' = 5x^4\frac{x^3 + 2}{x + 1} + (x^5+1)\frac{2x^3+3x^2-2}{(x+1)^2} \).

Step 1: Identify the Function Structure

The function is given as \( f(x) = (x^5 + 1) \frac{x^3 + 2}{x + 1} \). It is a product of two functions: \( u(x) = x^5 + 1 \) and \( v(x) = \frac{x^3 + 2}{x + 1} \). We'll differentiate it using the product rule.

Step 2: Apply the Product Rule

The product rule states that \( (uv)' = u'v + uv' \). So, we need to find the derivatives \( u'(x) \) and \( v'(x) \) and then apply the rule.

Step 3: Find \( u'(x) \)

Differentiate \( u(x) = x^5 + 1 \) with respect to \( x \). Using the power rule, \( u'(x) = 5x^4 \).

Step 4: Find \( v'(x) \) using Quotient Rule

The function \( v(x) = \frac{x^3 + 2}{x + 1} \) requires the quotient rule: if \( v(x) = \frac{a(x)}{b(x)} \), then \( v'(x) = \frac{a'(x)b(x) - a(x)b'(x)}{(b(x))^2} \). Here, \( a(x) = x^3 + 2 \) and \( b(x) = x + 1 \).- \( a'(x) = 3x^2 \)- \( b'(x) = 1 \)Thus, \( v'(x) = \frac{(3x^2)(x+1)-(x^3 + 2)(1)}{(x+1)^2} = \frac{3x^3 + 3x^2 - x^3 - 2}{(x+1)^2} = \frac{2x^3 + 3x^2 - 2}{(x+1)^2} \).

Step 5: Substitute into Product Rule Formula

Insert \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule:\[ (uv)' = (5x^4)\left(\frac{x^3 + 2}{x + 1}\right) + (x^5 + 1)\left(\frac{2x^3 + 3x^2 - 2}{(x+1)^2}\right) \].

Key Concepts

Product Rule

The product rule is a helpful method in calculus used to differentiate functions that are the product of two other functions. Imagine you have a function that looks like this: \( f(x) = g(x) \cdot h(x) \). The product rule tells us how to find the derivative, \( f'(x) \), of this function. It states that you take the derivative of \( g(x) \), multiply it by \( h(x) \), add the derivative of \( h(x) \), and multiply it by \( g(x) \). In formula form, this is written as:

• \( (g(x) \cdot h(x))' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \)

When you are given a problem with a function defined as a product of two component functions, like in this exercise with \( u(x) = x^5 + 1 \) and \( v(x) = \frac{x^3 + 2}{x + 1} \), you use the product rule to simplify differentiation. The product rule breaks down complex formulas into manageable sections by separately differentiating each part and combining them effectively. Practicing the product rule makes handling diverse functions easier.

Quotient Rule

The quotient rule is a formula used when you need to differentiate a quotient of two functions. It applies to situations where one function is divided by another, just like the function \( v(x) = \frac{x^3 + 2}{x + 1} \) in this exercise. To use the quotient rule, you need to know the derivatives of the top function \( a(x) \) and the bottom function \( b(x) \).The rule can be remembered as: the derivative of the top function times the bottom function, minus the top function times the derivative of the bottom function, all over the square of the bottom function:

• \( v'(x) = \frac{a'(x) \cdot b(x) - a(x) \cdot b'(x)}{(b(x))^2} \)

In our exercise, this means you need to differentiate \( a(x) = x^3 + 2 \) to get \( a'(x) = 3x^2 \), and differentiate \( b(x) = x+1 \) to get \( b'(x) = 1 \). Then apply these in the formula. Understanding and applying the quotient rule is crucial for handling division in calculus. It helps manage calculus problems by simplifying the derivative of a fraction into a straightforward computation.

Power Rule

The power rule is a fundamental tool in calculus for differentiating functions of the form \( x^n \), where \( n \) is any real number. It states that you bring the power down in front and then reduce the power by one. So, the derivative of \( x^5 \) becomes \( 5x^4 \). In this exercise, the function \( u(x) = x^5 + 1 \) is differentiated using the power rule.Here's how it breaks down:

• For \( x^5 \): \( \frac{d}{dx}[x^5] = 5x^4 \)
• The constant \( 1 \) disappears when differentiated, because the derivative of a constant is zero.

The power rule is particularly useful because it simplifies the process of taking derivatives, especially for polynomials. Mastering this rule makes it easier to tackle more complicated differentiation problems. Integrating the power rule into broader techniques, like the product and quotient rules, allows for effective and efficient calculus problem solving.