Y value for second derivative involving e^y

[Zealousy]

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:
$y'=\frac{-y}{x+9e^{y}}$

then, I found the second derivative to be:
$y''=\frac{y(1+9e^{y}*y')}{(x+9e^{y})^{2}}-\frac{y'}{x+9e^{y}}$

the website won't let me enter y' so I substituted it into the second derivative and then set x=0 and I got:
$y''=\frac{2y-y^{2}}{9e^{2y}}$

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!

 

[SammyS]

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:
$\displaystyle y'=\frac{-y}{x+9e^{y}}$

then, I found the second derivative to be:
$\displaystyle y''=\frac{y(1+9e^{y}\cdot y')}{(x+9e^{y})^{2}}-\frac{y'}{x+9e^{y}}$

the website won't let me enter y' so I substituted it into the second derivative and then set x=0 and I got:
$\displaystyle y''=\frac{2y-y^{2}}{9e^{2y}}$

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!

Plug x = 0 into $xy + 9e^y = 9e\,.$ Then solve for y . Use that in your expression for y''.

 

[dynamicsolo]

Homework Statement

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

...

Okay so first I found the first derivative using implicit differentiation and I got:
$y'=\frac{-y}{x+9e^{y}}$

then, I found the second derivative to be:
$y''=\frac{y(1+9e^{y}*y')}{(x+9e^{y})^{2}}-\frac{y'}{x+9e^{y}}$

the website won't let me enter y' so I substituted it into the second derivative and then set x=0 and I got:
$y''=\frac{2y-y^{2}}{9e^{2y}}$

Since you are asked for the value of $y''(0)$, you don't really need to fuss too much with an expression for $y''$.

First, are we agreed that y = 1 for x = 0 ? (I want to be careful, since you only have one attempt left on your computer-problem system.)

I agree with your result for $y'$. So we can say that $y'(0) = -\frac{1}{9e}$. Does that sound all right?

The first implicit differentiation of the original equation produced

$$y + xy' + 9 \cdot e^{y} \cdot y' = 0 ,$$

so a second implicit differentiation yields$$y' + ( y' + xy'' ) + ( 9 \cdot e^{y} \cdot y' \cdot y' + 9 \cdot e^{y} \cdot y'' ) = 0 \Rightarrow 2y' + xy'' + 9 \cdot e^{y} \cdot ( y' )^{2} + 9 \cdot e^{y} \cdot y'' = 0 .$$

You could now insert your results for $y(0)$ and $y'(0)$ , with x = 0 , then solve the resulting equation for $y''$.