Problem 91 - Calculus and its applications

Question

Differentiate. $$ y=\left(\frac{x^{2}-x-1}{x^{2}+1}\right)^{3} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 3 \left(\frac{x^2 - x - 1}{x^2 + 1}\right)^2 \cdot \frac{3x^2 + 1}{(x^2 + 1)^2} \).

Step 1: Identify the Derivative Rule(s) to Use

To differentiate the function \( y = \left(\frac{x^2 - x - 1}{x^2 + 1}\right)^3 \), observe it has a "function to a power" form, so you will need the chain rule. Additionally, since it involves a fraction, the quotient rule will also be necessary.

Step 2: Apply the outer Function's Derivative

Start with the derivative of the outer function. Using the chain rule, let \( u = \frac{x^2 - x - 1}{x^2 + 1} \). Then \( y = u^3 \) and \( \frac{dy}{du} = 3u^2 \). This guides you to differentiate the outer function as \( 3 \left(\frac{x^2 - x - 1}{x^2 + 1}\right)^2 \cdot \frac{du}{dx} \).

Step 3: Differentiate the Inner Function Using the Quotient Rule

For \( u = \frac{x^2 - x - 1}{x^2 + 1} \), the quotient rule gives you: \( \frac{d}{dx}[u] = \frac{(x^2 + 1)(2x - 1) - (x^2 - x - 1)(2x)}{(x^2 + 1)^2} \). Compute \( (x^2 + 1)(2x - 1) = 2x^3 + x^2 - 2x - 1 \) and \( (x^2 - x - 1)(2x) = 2x^3 - 2x^2 - 2x \).

Step 4: Simplify the Derivative of the Inner Function

Combine and simplify \( 2x^3 + x^2 - 2x - 1 \) and \( -(2x^3 - 2x^2 - 2x) \):\[ 2x^3 + x^2 - 2x - 1 - 2x^3 + 2x^2 + 2x = 3x^2 + 1 \]. Therefore, \( \frac{d}{dx}[u] = \frac{3x^2 + 1}{(x^2 + 1)^2} \).

Step 5: Combine Results Using the Chain Rule

Substitute \( \frac{du}{dx} = \frac{3x^2 + 1}{(x^2 + 1)^2} \) into the chain rule derivative to get:\[ \frac{dy}{dx} = 3 \left(\frac{x^2 - x - 1}{x^2 + 1}\right)^2 \cdot \frac{3x^2 + 1}{(x^2 + 1)^2}. \] Simplify the expression, if necessary.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in differential calculus used for differentiating composite functions. When you have one function nested inside another, the chain rule helps break down the process of differentiation into more manageable steps. In our example, we have a composite function:

• the outer function as a power function: \( y = u^3 \)
• the inner function, \( u = \frac{x^2 - x - 1}{x^2 + 1} \)

Here, the outer function is a simple power, and we can apply the derivative of \( u^3 \), which is \( 3u^2 \), then multiply it by the derivative of the inner function \( \frac{du}{dx} \). The chain rule formula tells us that the derivative of a composite function \( y = f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
In this case, \( f(u) = u^3 \) so \( f'(u) = 3u^2 \), and \( \frac{du}{dx} = \frac{(x^2 + 1)(2x - 1) - (x^2 - x - 1)(2x)}{(x^2 + 1)^2} \).
We then find the derivative of the entire function by combining these using the chain rule, simplifying the process of differentiating the complex expression.

Quotient Rule

The quotient rule is crucial when differentiating functions that are formed by the division of two other functions. In our exercise, the inner function \( u \) is a quotient \( u = \frac{x^2 - x - 1}{x^2 + 1} \). The quotient rule states that if you have a function \( h(x) = \frac{f(x)}{g(x)} \), its derivative is given by the formula:\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{ (g(x))^2 } \]Let's break it down for our specific case:

• \( f(x) = x^2 - x - 1 \)
• \( g(x) = x^2 + 1 \)
• The derivative \( f'(x) = 2x - 1 \)
• The derivative \( g'(x) = 2x \)

Using the quotient rule, we find:\[ \frac{d}{dx}[u] = \frac{(x^2 + 1)(2x - 1) - (x^2 - x - 1)(2x)}{(x^2 + 1)^2} \]After simplifying the expressions in the numerator, we end up with:\( 3x^2 + 1 \). This derivative helps us assess how the quotient changes with respect to \( x \). Integrating this into our chain rule operation enables us to find the derivative of the entire function.

Calculus Problem Solving

Differentiating complex functions often involves applying multiple rules and methodologies, such as the chain rule and quotient rule. Successful calculus problem-solving requires recognizing patterns in functions and discerning which rules are applicable.
Remember, the key steps in approaching differentiation problems like the one in the exercise include:

• Identifying the type of function you're differentiating. Is it a composite function? Does it involve a quotient?
• Choosing the correct differentiation rules. In this problem, it was both the chain rule and the quotient rule.
• Breaking down the components of the problem systematically. Address each part individually and integrate the results.
• Simplifying expressions where possible. This often makes both the math and final result more manageable.

Approaching problems step-by-step reduces errors and ensures clarity in finding derivatives of complex functions. These techniques form the bedrock of solving calculus problems efficiently and effectively.