Problem 53 - Brief Applied Calculus

Question

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

Step-by-Step Solution

Answer

Use the product and quotient rules to differentiate, and simplify the resulting expression.

Step 1: Apply the Product Rule

The function is a product of two expressions: \( f(x) = x^3 + 2 \) and \( g(x) = \frac{x^2 + 1}{x + 1} \). To differentiate a product of two functions, use the product rule: \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \).

Step 2: Differentiate the First Function

Differentiate \( f(x) = x^3 + 2 \) with respect to \( x \). The derivative \( f'(x) = 3x^2 \), as the constant derivative is zero.

Step 3: Differentiate the Second Function (Quotient Rule)

Differentiate \( g(x) = \frac{x^2 + 1}{x + 1} \) using the quotient rule: \[ \frac{d}{dx}[\frac{u}{v}] = \frac{u'v - uv'}{v^2} \]. \( u = x^2 + 1 \) and \( v = x + 1 \), so \( u' = 2x \) and \( v' = 1 \). Applying the quotient rule: \[ g'(x) = \frac{(2x)(x + 1) - (x^2 + 1)(1)}{(x + 1)^2} = \frac{2x^2 + 2x - x^2 - 1}{(x + 1)^2} = \frac{x^2 + 2x - 1}{(x + 1)^2} \].

Step 4: Substitute in the Product Rule

Substitute \( f'(x) = 3x^2 \), \( g(x) = \frac{x^2 + 1}{x + 1} \), \( f(x) = x^3 + 2 \), and \( g'(x) = \frac{x^2 + 2x - 1}{(x + 1)^2} \) into the product rule formula: \[ f'(x)g(x) + f(x)g'(x) = 3x^2 \left( \frac{x^2 + 1}{x + 1} \right) + (x^3 + 2) \left( \frac{x^2 + 2x - 1}{(x + 1)^2} \right) \].

Step 5: Simplify the Expression

To find the derivative, simplify the expression calculated in Step 4. First term: \( 3x^2 \cdot \frac{x^2 + 1}{x + 1} = \frac{3x^2(x^2 + 1)}{x + 1} \). Second term: \( (x^3 + 2) \cdot \frac{x^2 + 2x - 1}{(x + 1)^2} = \frac{(x^3 + 2)(x^2 + 2x - 1)}{(x + 1)^2} \). Combine and factorize where possible to simplify further if needed.

Key Concepts

Product Rule

The product rule is a crucial technique in calculus differentiation used to find the derivative of a product of two functions. If you have a function that's a product of two functions, say \( f(x) \) and \( g(x) \), the product rule enables you to differentiate this product. The rule is explicitly given by:

• \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)

By applying this rule, you systematically differentiate each function while leaving the other unchanged in each term of the sum, this means:

• First, take the derivative of \( f(x) \) and multiply it by \( g(x) \).
• Then, take the derivative of \( g(x) \) and multiply it by \( f(x) \).

Finally, add both results together to get the derivative of the product.

In our exercise, we have two functions \( f(x) = x^3 + 2 \) and \( g(x) = \frac{x^2 + 1}{x + 1} \). Using the product rule in step 4 requires substituting \( f'(x) = 3x^2 \) and \( g'(x) \) after it is calculated using the quotient rule.

This approach is essential since differentiating a product directly isn't simple, and this systematic method ensures that each part contributes correctly to the overall derivative.

Quotient Rule

When dealing with a division of two functions in calculus, like \( g(x) = \frac{x^2 + 1}{x + 1} \), the quotient rule is your go-to method for finding the derivative. It helps you when a function can be expressed as a ratio or fraction of two separate functions, \( u(x) \) and \( v(x) \). The quotient rule is stated as:

• \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2} \)

Here's how to break it down:

• \( u \) is your numerator function, and \( v \) is your denominator function.
• First, differentiate \( u \) to find \( u' \), and differentiate \( v \) to find \( v' \).
• Apply these derivatives to the formula above, ensuring to square \( v \) for the denominator.

In the provided exercise, you apply this rule to \( g(x) = \frac{x^2 + 1}{x + 1} \), with \( u = x^2 + 1 \) and \( v = x + 1 \). This leads to \( g'(x) \), a crucial step for then applying the product rule.

The quotient rule, though a bit more complex than the product rule, is straightforward with practice. Always remember to structure your differentiation with each part's contribution kept in mind.

Derivative Calculation

In the world of calculus, derivative calculation becomes an art of meticulous application of rules like the product and quotient rules. Derivatives represent the slope or rate of change of a function, making their calculation pivotal in understanding the function's behavior.

• Start by identifying which rules are applicable to your function. In our case, both the product and quotient rules are utilized.
• Clearly differentiate each part, whether they are products or quotients separately before combining results.

After applying these rules separately to our functions from the exercise, the next step involves simplifications. Often in derivative calculation:

• Simplifying expressions is key to achieving the most reduced and interpretable form of the derivative.
• Combine like terms and factorize expressions whenever possible to make the solution clear.

This methodical, layered approach ensures no differentiation step is overlooked and that each rule's application contributes to a correct and simplified result.

As shown in the step-by-step solution, after applying both the product and quotient rules, simplification becomes crucial. This not only aids in finalizing the derivative but also in checking the work for potential miscalculations. Remember, each derivative calculation can open new insights into the behavior of the function.