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## Homework Statement

If xy + 9e^y = 9e, find the value of y'' at the point where x = 0.

## Homework Equations

product rule

## The Attempt at a Solution

Okay so first I found the first derivative using implicit differentiation and I got:

[itex]y'=\frac{-y}{x+9e^{y}}[/itex]

then, I found the second derivative to be:

[itex]y''=\frac{y(1+9e^{y}*y')}{(x+9e^{y})^{2}}-\frac{y'}{x+9e^{y}}[/itex]

the website won't let me enter y' so I substituted it into the second derivative and then set x=0 and I got:

[itex]y''=\frac{2y-y^{2}}{9e^{2y}}[/itex]

I'm not quite sure if I'm doing this right so I'd like a second opinion. I only have one attempt left and I don't want to get it wrong!