# Y value for second derivative involving e^y

• Zealousy
In summary: Do you see how to do that?In summary, the value of y'' at the point where x = 0 is given by the expression y'' = \frac{2y-y^{2}}{9e^{2y}} where y = 1 and y' = -\frac{1}{9e}.

## Homework Statement

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

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!

Zealousy said:

## Homework Statement

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

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''.

Zealousy said:

## 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 $y''$.

## 1. What is the formula for finding the second derivative of e^y?

The formula for finding the second derivative of e^y is d^2/dx^2(e^y) = e^y.

## 2. Can the second derivative of e^y be negative?

Yes, the second derivative of e^y can be negative if the value of y is negative.

## 3. How does the value of y affect the second derivative of e^y?

The value of y does not directly affect the second derivative of e^y. However, the value of y can affect the first derivative, which in turn can affect the second derivative.

## 4. Is the second derivative of e^y always positive?

No, the second derivative of e^y is not always positive. It can be positive, negative, or zero, depending on the value of y.

## 5. Can the second derivative of e^y be used in real-life applications?

Yes, the second derivative of e^y has many real-life applications in fields such as physics, chemistry, and engineering. It can be used to model exponential growth and decay, among other things.