## Calculus: Early Transcendentals 8th Edition

$f'(x) = -\frac{x^{2}-1}{(x^{2}-1)^{2}}$ $f''(x) = \frac{2x(x^{2} + 3)}{(x^{2}-1)^{3}}$
Find $f'(x)$ $f'(x) = \frac{\frac{d(x)}{dx}(x^2-1)-\frac{d(x^{2}-1)}{dx}(x)}{(x^{2}-1)^{2}}$ $f'(x) = \frac{1(x^{2}-1)-2x(x)}{(x^{2}-1)^{2}}$ $f'(x) = -\frac{x^{2}-1}{(x^{2}-1)^{2}}$ $f'(x) = \frac{x^2 - 1 - 2x^2}{(x^{2}-1)^{2}}$ $f'(x) = -\frac{x^{2}-1}{(x^{2}-1)^{2}}$ Find $f''(x)$ $f''(x) = \frac{\frac{d(-x^2 -1)^{2}}{dx}(x^2-1)^{2}-\frac{d(x^{2}-1)^{2}}{dx}(-x^2 - 1)}{((x^{2}-1)^{2})^{2}}$ $f''(x) = \frac{(-2x)(x^2-1)^{2} - 4x(x^2 -1)(-x^2 -1)}{(x^2 - 1)^{4}}$ Factor $2x(x^2 -1)$: $f''(x) = \frac{2x(x^2 -1)[-(x^2 -1) -2(-x^2 -1)]}{(x^2 - 1)^{4}}$ $f''(x) = \frac{2x(x^2 -1)[-x^2 + 1 + 2x^2 + 2]}{(x^2 - 1)^{4}}$ $f''(x) = \frac{2x(x^2 -1)[x^2 + 3]}{(x^2 - 1)^{4}}$ Cancel $(x^2 -1)$ from the numerator: $f''(x) = \frac{2x(x^{2} + 3)}{(x^{2}-1)^{3}}$