Problem 62 - Calculus and its applications

Question

Consider $$ f(x)=\frac{x^{2}}{(1+x)^{5}} $$ a) Find \(f^{\prime}(x)\) using the Quotient Rule and the Extended Power Rule. b) Note that \(f(x)=x^{2}(1+x)^{-5}\). Find \(f^{\prime}(x)\) using the Product Rule and the Extended Power Rule. c) Compare your answers to parts (a) and (b).

Step-by-Step Solution

Answer

Both methods yield the derivative: \(f'(x) = \frac{2x - 3x^2}{(1+x)^6}\).

Step 1: Apply the Quotient Rule

To find the derivative of the function \(f(x) = \frac{x^2}{(1+x)^5}\), we'll apply the quotient rule. The quotient rule states that for a function \(u/v\), its derivative is \(\frac{{u'v - uv'}}{v^2}\). Here, \(u = x^2\) and \(v = (1+x)^5\). First, find the derivatives: \(u' = 2x\) and \(v' = 5(1+x)^4\).

Step 2: Calculate the Derivative Using the Quotient Rule

Substitute the derivatives into the quotient rule formula:\[f'(x) = \frac{(2x)(1+x)^5 - (x^2)(5(1+x)^4)}{((1+x)^5)^2}\]Simplify the expression:\[f'(x) = \frac{2x(1+x)^5 - 5x^2(1+x)^4}{(1+x)^{10}}\]

Step 3: Simplify the Derivative Expression

Factor \((1+x)^4\) from the numerator:\[f'(x) = \frac{(1+x)^4(2x(1+x) - 5x^2)}{(1+x)^{10}}\]Simplify further:\[f'(x) = \frac{(1+x)^4(2x + 2x^2 - 5x^2)}{(1+x)^{10}}\]\[f'(x) = \frac{(1+x)^4(2x - 3x^2)}{(1+x)^{10}}\]

Step 4: Final Simplification for Part (a)

Further simplify the expression:\[f'(x) = \frac{(2x - 3x^2)}{(1+x)^6}\]This is \(f'(x)\) using the quotient rule.

Step 5: Rewrite the Function for the Product Rule

Given \(f(x) = x^2(1+x)^{-5}\), we can apply the product rule, which states that for a function \(u\cdot v\), its derivative is \(u'v + uv'\). Here, \(u = x^2\) and \(v = (1+x)^{-5}\). First, find the derivatives: \(u' = 2x\) and \(v' = -5(1+x)^{-6}\).

Step 6: Calculate the Derivative Using the Product Rule

Substitute the derivatives into the product rule formula:\[f'(x) = 2x(1+x)^{-5} + x^2(-5)(1+x)^{-6}\]Simplify:\[f'(x) = 2x(1+x)^{-5} - 5x^2(1+x)^{-6}\]

Step 7: Simplify the Product Rule Result

Combine the terms over a common denominator:\[f'(x) = \frac{2x(1+x) - 5x^2}{(1+x)^6}\]This matches the form: \(f'(x) = \frac{2x - 3x^2}{(1+x)^6}\).

Step 8: Compare Results of Part (a) and (b)

Both the derivative found using the quotient rule and the product rule result in:\[f'(x) = \frac{2x - 3x^2}{(1+x)^6}\]This confirms that the results from both methods are equivalent.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental tool in Calculus used for finding the derivative of a quotient of two functions. Whenever you have a function in the form of \(u/v\), where both \(u\) and \(v\) are differentiable functions, the Quotient Rule helps determine the derivative.

The formula for the Quotient Rule is given by:

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

Here, \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.

Applying this to our original example, where \( u = x^2 \) and \( v = (1+x)^5 \), involves:

• Finding \( u' = 2x \)
• Finding \( v' = 5(1+x)^4 \)
• Substituting these into the Quotient Rule formula

This results in the derivative expression:

• \( f'(x) = \frac{(2x)(1+x)^5 - (x^2)(5(1+x)^4)}{((1+x)^5)^2} \)

After simplification, you achieve the simplified form of the derivative:

• \( f'(x) = \frac{(2x - 3x^2)}{(1+x)^6} \)

This process shows how the Quotient Rule enables us to handle complex fraction derivatives with ease.

Product Rule

The Product Rule is crucial when differentiating the product of two functions. It tells us how to differentiate two multiplied functions where each function is individually differentiable.

If you have a function in the form of \( u \cdot v \), then the Product Rule states that:

• \( \frac{d}{dx}(u \cdot v) = u'v + uv' \)

Applying this rule to find the derivative of the function \( f(x) = x^2(1+x)^{-5} \):

Let:

• \( u = x^2 \) with \( u' = 2x \)
• \( v = (1+x)^{-5} \) with \( v' = -5(1+x)^{-6} \)

By substituting back, the product rule formula gives:

• \( f'(x) = 2x(1+x)^{-5} + x^2(-5)(1+x)^{-6} \)

Simplifying this expression:

• \( f'(x) = \frac{2x(1+x) - 5x^2}{(1+x)^6} \)

Which also simplifies to:

• \( f'(x) = \frac{2x - 3x^2}{(1+x)^6} \)

Thus, the Product Rule provides another way to find the same derivative, showcasing its flexibility.

Derivative Calculation

Derivative Calculation is a key process in Calculus that allows us to find the rate at which a function is changing. When calculating derivatives, several rules, like the Quotient Rule and Product Rule, are applied depending on the structure of the function.

In our example, we calculated the derivative of the function \( f(x) = \frac{x^2}{(1+x)^5} \) using both the Quotient Rule and Product Rule with the Extended Power Rule.

The outcome from either method is:

• \( f'(x) = \frac{2x - 3x^2}{(1+x)^6} \)

This demonstrates that different differentiation techniques can lead us to the same result. The Extended Power Rule assists particularly with terms like \( (1+x)^{-5} \), making differentiation of power functions straightforward.

The ability to use different methods not only confirms the reliability of Calculus rules but also builds confidence in executing complex derivative calculations effectively. Understanding derivative calculation transforms how we analyze function behaviors across many fields of science, engineering, and economics.