Homework Help: Derivatives, chain rule 6

1. Nov 10, 2013

physics604

1. The problem statement, all variables and given/known data

$$xy+e^y=e$$ Find the value of y'' at the point where x=0.

2. Relevant equations

Chain rule

3. The attempt at a solution

I find y' first.

$$x\frac{dy}{dx}+y+e^y\frac{dy}{dx}=0$$ $$(x+e^y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=\frac{-y}{x+e^y}$$

Then I find y''.

$$\frac{(x+e^y)(-\frac{dy}{dx})-(-y)(1+e^y\frac{dy}{dx})}{(x+e^y)^2}$$ $$\frac{(x+e^y)(\frac{y}{x+e^y})-(-y-ye^y(\frac{-y}{x+e^y}))}{(x+e^y)^2}$$ $$\frac{y-(-y+\frac{y^2e^y}{x+e^y})}{(x+e^y)^2}$$ $$\frac{y+y-\frac{y^2e^y}{x+e^y}}{(x+e^y)^2}$$ $$(2y-\frac{y^2e^y}{x+e^y})(\frac{1}{(x+e^y)^2})$$ $$\frac{2y(x+e^y)-y^2e^y}{(x+e^y)^3}$$ $$\frac{2xy+2ye^y-y^2e^y}{(x+e^y)^3}$$

x=0 so plugging that in I get

$$\frac{ye^y(2-y)}{e^{3y}}=\frac{y(2-y)}{e^{2y}}$$

The answer is $$\frac{1}{e^2}$$ What did I do wrong?

Any help is much appreciated.

2. Nov 10, 2013

Dick

You aren't finished. Same issue as in the last problem. Did you try using your function definition to simplify that?

3. Nov 10, 2013

physics604

I've tried plugging in $$e^y=e-xy$$ but I don't get anywhere.

$$\frac{y(2-y)}{e^{2y}}=\frac{y(2-y)}{(e-xy)^2}$$

x=0 so

$$\frac{y(2-y)}{e^2}$$

4. Nov 10, 2013

Dick

Put x=0 into your original equation. Can you conclude anything about the value of y when x=0?

5. Nov 10, 2013

physics604

What do you mean?

If I plug x=0 into my original equation I get $$e^y=e-(0)y$$ so $$e^y=e$$ Or I can rearrange the equation to get $$y=\frac{e-e^y}{x}=\frac{e-e^y}{0}=undefined$$

6. Nov 10, 2013

Dick

You should be able to figure out what y is when x=0 from e^y=e. What is it?

7. Nov 10, 2013

physics604

y=1. Okay, thanks!