Question

Seja f(x) = 2x+1 e g(x) = $\sqrt[3]{x^{2} }$
Então, calcule (gof)'(x):

Answer 1

$g(x)=\sqrt[3]{x^2}\\\\(g\circ f)(x)=\sqrt[3]{(2x+1)^2}\\\\(g\circ f)(x)=(2x+1)^{\frac{2}{3}}\\\\(g\circ f)'(x)=\frac{2}{3}(2x+1)^{\frac{2}{3}-\frac{3}{3}}(2)\\\\(g\circ f)'(x)=\frac{4}{3}(2x+1)^{-\frac{1}{3}}\\\\(g\circ f)'(x)=\frac{4}{3\sqrt[3]{2x+1}}$