Question

Помогите найти производную функции: f(x) = ln (x^2+4)/(x^2-1); f'(2)

Answer 1

$f'(x)= \frac{x^2-1}{x^2+4} * \frac{2x(x^2-1)-2x(x^2+4)}{(x^2-1)^2} = \frac{-10x}{(x^2+4)(x^2-1)}$
$f'(2)= \frac{-10*2}{(2^2+4)(2^2-1)} =- \frac{20}{24}=- \frac{5}{6}$