Problem 26 - Single Variable Calculus

Question

Find the derivative of each function. \(f(x)=\frac{x}{x^{2}-4}-\frac{x-1}{x^{2}+4}\)

Step-by-Step Solution

Answer

The derivative of \(f(x)=\frac{x}{x^{2}-4}-\frac{x-1}{x^{2}+4}\) is: \(f'(x) = \frac{-x^2 - 4}{(x^2 - 4)^2} - \frac{-x^2 + 2x + 4}{(x^2 + 4)^2}\)

Step 1: Define the functions

Define the two functions given in the problem: 1. \(f_1(x) = \frac{x}{x^2 - 4}\) 2. \(f_2(x) = \frac{x - 1}{x^2 + 4}\) Our goal is to find the derivative of \(f(x) = f_1(x) - f_2(x)\).

Step 2: Find the Derivative of \(f_1(x)\)

Applying the quotient rule to \(f_1(x) = \frac{x}{x^2 - 4}\), we get: \(f_1'(x) = \frac{1 \cdot (x^2 - 4) - x \cdot 2x}{(x^2 - 4)^2}\) Simplify the expression: \(f_1'(x) = \frac{x^2 - 4 - 2x^2}{(x^2 - 4)^2}\) \(f_1'(x) = \frac{-x^2 - 4}{(x^2 - 4)^2}\)

Step 3: Find the Derivative of \(f_2(x)\)

Applying the quotient rule to \(f_2(x) = \frac{x - 1}{x^2 + 4}\), we get: \(f_2'(x) = \frac{1 \cdot (x^2 + 4) - (x - 1) \cdot 2x}{(x^2 + 4)^2}\) Simplify the expression: \(f_2'(x) = \frac{x^2 + 4 - 2x^2 + 2x}{(x^2 + 4)^2}\) \(f_2'(x) = \frac{-x^2 + 2x + 4}{(x^2 + 4)^2}\)

Step 4: Combine Both Derivatives

Now that we have the derivatives of both \(f_1(x)\) and \(f_2(x)\), we can compute the derivative of their difference using the subtraction rule. Recall that \(f(x) = f_1(x) - f_2(x)\), so its derivative is given by: \(f'(x) = f_1'(x) - f_2'(x)\) Now substitute the derivatives we found in Steps 2 and 3: \(f'(x) = \frac{-x^2 - 4}{(x^2 - 4)^2} - \frac{-x^2 + 2x + 4}{(x^2 + 4)^2}\)

Step 5: Final Answer

We have computed the derivative of the given function as: \(f'(x) = \frac{-x^2 - 4}{(x^2 - 4)^2} - \frac{-x^2 + 2x + 4}{(x^2 + 4)^2}\) This is the final answer.