If $y=\ln(\frac{1}{3}(1+e^{-2x}))$, show that $\frac{dy}{dx}=\frac{2}{3}(e^{-y}-3)$. I don't understand why there is only the $y$ variable in the derivative.
I have differentiated it directly and I got $\frac{dy}{dx}=\frac{-2e^{-2x}}{1+e^{-2x}}$, but I don't really see how I can get to the final answer. Any guidance please?
 A: $$y=\ln(\frac{1}{3}(1+e^{-2x}))\to e^y=\frac{1}{3}+\frac{1}{3}e^{-2x}\to y'e^y=-\frac{2}{3}e^{-2x}=-2(e^y-\frac{1}{3})\to \\ y'=\frac{dy}{dx}=-2(1-\frac{1}{3}e^{-y})=\frac{2}{3}(e^{-y}-3)$$
A: This question asks you to prove the the given function $y(x)$ satisfies the given differential equation. It's not really a "find something" kind of exercise, but rather a "prove something" kind of exercise.
You've done half of the work already: you found the left-hand side $\frac{dy}{dx}$ of the equation. Now the other half is to evaluate the right-hand side $\frac{2}{3}(e^{-y}-3)$ of the equation: plug in the given formula for $y$ into this expression and then simplify it using properties of exponents and logarithms.
After you do that, hopefully you will see that the expressions you obtained for both sides are equal to each other — which is exactly the point of this exercise.
A: $$y'=\frac 1 {(\frac{1}{3}(1+e^{-2x}))}\frac{1}{3}(-2e^{-2x})=\frac {-2e^{-2x} }{(1+e^{-2x})}$$
But the first equality gives you a relation between y and x
$$e^{-2x}=3e^y-1$$
$$y'=\frac{2}{3}(e^{-y}-3)$$
A: You have $$\frac{dy}{dx}=\frac{-2e^{-2x}}{1+e^{-2x}}$$
From $y=\ln\frac13(1+e^{-2x})$, we have that $$3e^y-1=e^{-2x}$$So substituting this into your expression, $$\frac{dy}{dx}=\frac{-2(3e^y-1)}{1+3e^y-1}=\frac23\left(\frac{1-3e^y}{e^y}\right)=\frac23\left(e^{-y}-3\right)$$

An alternative method is $$x=-\frac12\ln(3e^y-1)\\\frac{dx}{dy}=-\frac12\frac{3e^y}{3e^y-1}\\\frac{dy}{dx}=\frac23(-3+e^{-y})$$
