Question

determinar a derivada da função x²+8/x³

Answer 1

Regra do quociente

$\large\boxed{(\frac{u}{v})'=\frac{u'v-uv'}{v^2}}$

Então, temos que:

$(\frac{x^2+8}{x^3})'=\frac{2x(x^3)-(x^2+8)(3x^2)}{(x^3)^{2}}\\\\ (\frac{x^2+8}{x^3})'=\frac{2x^4-3x^4+24x^2}{x^6}\\\\ (\frac{x^2+8}{x^3})'=\frac{x^2(2x^2-3x^2+24)}{(x^6)}\\\\ \large\boxed{(\frac{x^2+8}{x^3})'=\frac{-x^2+24}{x^4}}$