Question

How to find the derivative???

Answer 1

Answer:

$y'= \dfrac{2}{3}e^{2x + e^{2x}/3}$

Step-by-step explanation:

We have $e^{2x} = \ln(y^3)$ and we want the implicit differentiation $\dfrac{dy}{dx}f(x) = y'$

Therefore, we want to separate the $y$ from the equation.

$\ln(y^3) =e^{2x} \implies 3 \ln(y) = e^{2x} \implies \ln(y) = \dfrac{e^{2x} }{3}$

$\implies e^{\ln(y)} = e^{e^{2x}/3} \implies \boxed{y = e^{e^{2x}/3}}$

Now we can calculate the derivative. By the chain rule,

$y' = \dfrac{d}{dx}\left(e^{e^{2x}/3}\right) = e^{e^{2x}/3}\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right)$

Now

$\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right) =\dfrac{1}{3}\dfrac{d}{dx}\left(e^{2x}\right) = \dfrac{2}{3}e^{2x}$

Therefore

$y' = e^{e^{2x}/3}\cdot \dfrac{2}{3}e^{2x} = \dfrac{2}{3}e^{2x + e^{2x}/3}$