Questions: 1. f(x)=(x^2-1)/(x^2+x+4) find f'(x)

Solution Steps

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{x^2 - 1}{x^2 + x + 4} \), we can use the quotient rule. The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then its derivative is given by:

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]

Here, \( g(x) = x^2 - 1 \) and \( h(x) = x^2 + x + 4 \). We need to find \( g'(x) \) and \( h'(x) \) and then apply the quotient rule.

Given the function \( f(x) = \frac{x^2 - 1}{x^2 + x + 4} \), we identify the numerator \( g(x) = x^2 - 1 \) and the denominator \( h(x) = x^2 + x + 4 \).

We need to find the derivatives \( g'(x) \) and \( h'(x) \): \[ g'(x) = \frac{d}{dx}(x^2 - 1) = 2x \] \[ h'(x) = \frac{d}{dx}(x^2 + x + 4) = 2x + 1 \]

Using the quotient rule: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \] Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \): \[ f'(x) = \frac{(2x)(x^2 + x + 4) - (x^2 - 1)(2x + 1)}{(x^2 + x + 4)^2} \]

Simplify the numerator: \[ (2x)(x^2 + x + 4) = 2x^3 + 2x^2 + 8x \] \[ (x^2 - 1)(2x + 1) = 2x^3 + x^2 - 2x - 1 \] Combine these results: \[ 2x^3 + 2x^2 + 8x - (2x^3 + x^2 - 2x - 1) = 2x^2 + 8x - x^2 + 2x + 1 = x^2 + 10x + 1 \]

Thus, the derivative is: \[ f'(x) = \frac{x^2 + 10x + 1}{(x^2 + x + 4)^2} \]

Final Answer