Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{2}{5 x^{2}-1} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-20x}{(5x^2 - 1)^2} \).

Step 1: Identify the Function and Rules

The function given is \( y = \frac{2}{5x^2 - 1} \). We need to find the derivative \( D_{x} y \) using differentiation rules. Notice it's a quotient with a constant numerator, which suggests using the derivative rule for a rational function, specifically the chain rule and power rule.

Step 2: Rewrite the Function for Differentiation

Express the function such that the variable \( x \) is in the numerator: \( y = 2 \cdot (5x^2 - 1)^{-1} \). This will allow us to use the chain rule and power rule effectively.

Step 3: Apply the Chain Rule and Power Rule

To differentiate \( y = 2 \cdot (5x^2 - 1)^{-1} \), apply the chain rule and power rule. The outer function is \((u)^{-1}\) where \( u = 5x^2 - 1 \), and has a derivative of \(-u^{-2}\). Multiply by the derivative of \(u\), which is \(10x\).

Step 4: Compute the Derivative

Differentiate using the above steps: \[D_{x} y = 2 \cdot (-1) \cdot (5x^2 - 1)^{-2} \cdot 10x = -20x(5x^2 - 1)^{-2} \]Simplify this expression: \[D_{x} y = \frac{-20x}{(5x^2 - 1)^2}\]

Step 5: Summarize the Result

The final derivative of the function \( y = \frac{2}{5x^2 - 1} \) with respect to \( x \) is \( \frac{-20x}{(5x^2 - 1)^2} \). Ensure that all steps were followed according to derivative rules, combining power and chain rules properly.

Key Concepts

Chain Rule

The chain rule is a foundational concept in calculus used to differentiate composite functions. A composite function is essentially a function nested inside another function. The chain rule allows you to break these functions into their simpler components to find the derivative.

Imagine you have a function \( y \) that depends on a function \( u \), and \( u \) depends on \( x \). Mathematically, this is expressed as \( y = f(u) \) and \( u = g(x) \).

• First, find the derivative of \( f(u) \) with respect to \( u \).
• Then, find the derivative of \( u \) with respect to \( x \).
• Finally, multiply these derivatives together.

This process makes differentiating complex functions manageable. In the example of \( y = 2(5x^2 - 1)^{-1} \), the outer function is \( u^{-1} \), where \( u = 5x^2 - 1 \). Differentiating the outer function and multiplying by the derivative of \( u \) provides the result.

Power Rule

The power rule is one of the simplest and most commonly used rules in differentiation. It's a method to find the derivative of expressions where a variable is raised to a power. Suppose you have a basic power function \( y = x^n \). The power rule states:

\[ D_x y = nx^{n-1} \]

What this means is you bring the exponent down, multiply it by the base raised to one less than the original exponent.

• If \( y = x^3 \), the derivative is \( D_x y = 3x^2 \).
• If \( y = x^{-1} \), then \( D_x y = -1x^{-2} \).

In the original problem, the power rule is applied to the re-expressed function \( (5x^2 - 1)^{-1} \). By bringing down the power and reducing the original exponent by one, and then multiplying by the derivative of \( u \), you get the derivative.

Rational Function Differentiation

Rational functions are quotients of two polynomials, such as \( y = \frac{2}{5x^2 - 1} \). Differentiating rational functions usually involves rules such as the quotient rule and chain rule. However, when the numerator is a constant, the chain rule combined with the power rule can be much simpler.

For these functions, first rewrite the function so the variable is in the numerator by expressing it as a negative power. Then apply the chain rule and power rule in steps:

• Identify the function's structure and re-express if necessary.
• Apply the power rule to the new form of the function.
• Apply the chain rule to handle the inner function’s derivative.

The solution was re-expressed as \( 2 \cdot (5x^2 - 1)^{-1} \), turning it into a manageable expression for applying these differentiation rules.