Question

Socorro! Como derivar essa função?
$y= e^{2x}( \sqrt[3]{ 3x^{2} } + \frac{1}{ 4x^{- \frac{1}{2} } })^4$

Answer 1

Oi Anna 
Use a regra do produto, pois temos uma função exponencial multiplicando uma algébrica:

$y=e^{2x}( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 \\ \\ y= \frac{d}{dx} (e^{2x})( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 +e^{2x} \frac{d}{dx} ( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 \\ \\ y'=2e^{2x}( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 +e^{2x}4(\sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }})^3.((3x^2)^{ \frac{1}{3} + \frac{x^{ \frac{1}{2} }}{4} })'$
$y'=2e^{2x}( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 +e^{2x}4(\sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }})^3.( \frac{1}{3} (3x^2)^{ -\frac{2}{3}}6x+ \frac{1}{2}\frac{x^{ -\frac{1}{2} }}{4} }) \\ \\ y'=2e^{2x}( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 +e^{2x}4(\sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }})^3.( \frac{2x}{(3x^2)^{ \frac{2}{3}} }+ \frac{1}{8x^{ \frac{1}{2} }} })$
$\boxed{ y'=2e^{2x}( \sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }} )^4 +e^{2x}(\sqrt[3]{3x^2}+ \frac{1}{4x^{- \frac{1}{2} }})^3.( \frac{8x}{ \sqrt[3]{(3x^2)^2} }+ \frac{1}{2 \sqrt{x} } })}$

Espero que goste