Answer:
[tex]y'= \dfrac{2}{3}e^{2x + e^{2x}/3}[/tex]
Step-by-step explanation:
We have [tex]e^{2x} = \ln(y^3)[/tex] and we want the implicit differentiation [tex]\dfrac{dy}{dx}f(x) = y'[/tex]
Therefore, we want to separate the [tex]y[/tex] from the equation.
[tex]\ln(y^3) =e^{2x} \implies 3 \ln(y) = e^{2x} \implies \ln(y) = \dfrac{e^{2x} }{3}[/tex]
[tex]\implies e^{\ln(y)} = e^{e^{2x}/3} \implies \boxed{y = e^{e^{2x}/3}}[/tex]
Now we can calculate the derivative. By the chain rule,
[tex]y' = \dfrac{d}{dx}\left(e^{e^{2x}/3}\right) = e^{e^{2x}/3}\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right)[/tex]
Now
[tex]\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right) =\dfrac{1}{3}\dfrac{d}{dx}\left(e^{2x}\right) = \dfrac{2}{3}e^{2x}[/tex]
Therefore
[tex]y' = e^{e^{2x}/3}\cdot \dfrac{2}{3}e^{2x} = \dfrac{2}{3}e^{2x + e^{2x}/3}[/tex]
