Question

Derivada de
$y= \frac{1}{(4x^3+5x^2+4)^5}$

Answer 1

$y = \frac{1}{(4x^3+5x^2 +4)^5}$

vamos passar o denominador para cima trocando o sinal:

$\\ y = (4x^3+5x^2+4)^-^5 \\ \\ y' = [(4x^3+5x^2+4)^-^5]'*(4x^3+5x^2+4)' \\ \\ y' = -5(4x^3+5x^2+4)^-^5^-^1*(3*4x^3^-^1+2*5x^2^-^1+0) \\ \\ y' = -5(4x^3+5x^2+4)^-^6(12x^2+10x) \\ \\ y' = \frac{-5}{(4x^3+5x^2+4)^6} *(12x^2+10x) \\ \\ y' = \frac{-5*12x^2-5*10x}{(4x^3+5x^2+4)^6} \\ \\ y' = \frac{-60x^2-50x}{(4x^3+5x^2+4)^6} \\ \\ y' = \frac{10(-6x^2-5x)}{(4x^3+5x^2+4)^6}$