Помогите решить 3 номер

Ответ:

$f'(x) = -2sin(x) +\frac{2}{(x+4)^2} - 18x^2$

Пошаговое объяснение:

$f(x) = 2cos(x) - \frac{x-6}{x+4}-6x^3\\f'(x) = (2cos(x))' - (\frac{x-6}{x+4})' - (6x^3)'\\f'(x) = -2sin(x) - \frac{(x-6)'(x+4) - (x-6)(x+4)'}{(x+4)^2} - 18x^2\\f'(x) = -2sin(x) - \frac{x+4-(x-6)}{(x+4)^2}-18x^2\\f'(x) = -2sin(x) +\frac{2}{(x+4)^2} - 18x^2$

Так же можно использовать данный сайт

https://www.wolframalpha.com/input/?i=derivative&assumption=%7B%22F%22,+%22Derivative%22,+%22derivativefunction%22%7D+-%3E%222*cos(x)+-+(x-6)%2F(x%2B4)+-+6x%5E3%22&assumption=%7B%22FVarOpt%22,+%221%22%7D+-%3E+%7B%7B%7D,+%7B%7B%7B%22Derivative%22,+%22derivativevariable%22%7D%7D%7D,+%7B%7D%7D&assumption=%7B%22C%22,+%22derivative%22%7D+-%3E+%7B%22Calculator%22,+%22dflt%22%7D