Question

Derivada de
$y= \frac{ \sqrt[3]{t} }{3t-5}$

Answer 1

$\\ y = \frac{ \sqrt[3]{T} }{3T-5} \\ \\ y = \frac{T^1^/^3}{3T-5} \\ \\ y' = \frac{(T^1^/^3)'*(3T-5)-T^1^/^3*(3T-5)'}{(3T-5)^2} \\ \\ y' = \frac{ \frac{1}{3}*T^1^/^3^-^1(3T-5)-T^1^/^3(3-0) }{(3T-5)^2} \\ \\ y' = \frac{ \frac{1}{3}*T^-^2^/^3(3T-5)-3T^1^/^3 }{(3T-5)^2} \\ \\ y' = \frac{ \frac{1*(3T-5)}{3T^2^/^3} -3T^1^/^3}{(3T-5)^2} \\ \\ y' = \frac{ \frac{(3t-5)-(3T^2^/^3)*(-3T^1^/^3)}{3T^2^/^3} }{(3T-5)^2}$

$\\ y' = \frac{ \frac{(3t-5)-(3T^2^/^3)*(-3T^1^/^3)}{3T^2^/^3} }{(3T-5)^2} \\ \\ y' = \frac{3T-5-9T^2^/^3^+^1^/^3}{3T^2^/^3(3T-5)^2} \\ \\ y' = \frac{3T-5-9T^3^/^3}{3 \sqrt[3]{T^2}(3T-5)^2 } \\ \\ y' = \frac{3T-5-9T}{3 \sqrt[3]{T^2}(3T-5)^2 } \\ \\ y' = \frac{-5-6T}{3 \sqrt[3]{T^2} (#T-5)^2}$