Implicit differentiation with e

Question

I am trying to find $\frac{dy}{dx}$ of $$e^{2y}+2e^x = 3$$ I am able to get as far as differentiating both sides of the equation, but then I struggle in the algebra to solve for y. Can someone hold my hand and help me finish this?

So far, I know:

$$2y(e^{2y'})+2e^x = 0$$ $$2ye^{2y'} = -2e^x$$

then I get stuck

Answer 1

Step 1: Taking derivative both sides, and remember to use the Chain Rule:

$2y'e^{2y} + 2e^x = 0$

Step 2: Solve for $y'$.

$y' = \dfrac{-e^x}{e^{2y}}$

Step 3: Substitute $3-2e^x$ for $e^{2y}$

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

Answer 2

Do check your usage of the chain rule. It should be

$$\frac{d}{dx} e^{2y} = e^{2y}\frac{d}{dx}(2y)= 2y'e^{2y}$$

Answer 3

This is one problem where I prefer to use the notation of Leibniz ($\frac{dy}{dx}$) over that of LaGrange ($y^\prime$), initially. I would set it up as follows.

$$\dfrac{de^{2y}}{d2y}\cdot \dfrac{d2y}{dy}\cdot \dfrac{dy}{dx}+\dfrac{d2e^x}{de^x}\cdot\dfrac{de^x}{dx}=\dfrac{d3}{dx}$$ Evaluating and rewriting $\frac{dy}{dx}$ as $y^\prime$ we have $$e^{2y}\cdot 2\cdot y^\prime+2\cdot e^x=0$$ $$2y^\prime e^{2y}+2e^x=0$$ Now just solve for $y^\prime$.