Question

Derivada de
$y=3x_{3}^2-4x_{4}^1 -2$

Answer 1

$\\ y = 3x^2^/^3-4x^1^/^4 -2 \\ \\ y' = 3 *\frac{2}{3} x^2^/^3^-^1-4 *\frac{1}{4} x^1^/^4^-^1+0 \\ \\ y' = 2x^-^1^/^3 -x^-^3^/^4 \\ \\ y' = \frac{2x}{x^1^/^3} - \frac{1}{x^3^/^4} \\ \\ y' = \frac{2x*x^3^/^4-x^1^/^3}{x^1^/^3*x^3^/^4} \\ \\ y' = \frac{2x^1^+^3^/^4-x^1^/^3}{x^1^/^3^+^3^/^4} \\ \\ y' = \frac{2x^7^/^4-x^1^/^3}{x^1^3^/^1^2} \\ \\ y' = \frac{2 \sqrt[4]{x^7} - \sqrt[3]{x} }{ \sqrt[12]{x^1^3} }$

Answer 2

$\sf \displaystyle y=3x\frac{2}{3}-4x\frac{1}{4}-2\\\\\\\frac{d}{dx}\left(3x\frac{2}{3}-4x\frac{1}{4}-2\right)\\\\\\\bullet{Aplicar\:a\:regra\:da\:soma/diferenca}:\to \quad \left(f\pm g\right)'=f\:'\pm g'\\\\\\\frac{d}{dx}\left(3x\frac{2}{3}\right)=2\\\\\\\frac{d}{dx}\left(4x\frac{1}{4}\right)=1\\\\\\\frac{d}{dx}(2)=0\\\\\\=2-1-0\\\\\\\to \boxed{\sf =1}$