# Second derivative in terms of x and y?

## Homework Statement

Find y" in terms of x and y:

y^2 + 2y = 2x + 1

N/A

## The Attempt at a Solution

I found the first derivative:

y^2 + 2y = 2x + 1
2yy'+2y'=2
2y'.(y+1)=2
y'=2/2(y+1)
y'=1/(y+1)

But I'm having trouble moving on from there.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find y" in terms of x and y:

y^2 + 2y = 2x + 1

N/A

## The Attempt at a Solution

I found the first derivative:

y^2 + 2y = 2x + 1
2yy'+2y'=2
2y'.(y+1)=2
y'=2/2(y+1)
y'=1/(y+1)

But I'm having trouble moving on from there.
Now, take the derivative of y'.

Sure it will have y' in it, but then substitute the result you have for y' into that.

Now, take the derivative of y'.

Sure it will have y' in it, but then substitute the result you have for y' into that.

alrighty so...

y" = derivative of 1/y+1

= (1)(y+1)^-1 ... then use product rule

= 0 + (-1(y+1)^-2)y' ... then plug in y'

= - [1/(y+1)]/(y+1)^2 ... combine

y" = - 1/(y+1)^3 final answer

......

I think I did it right. Does this satisfy "in terms of x and y?"

SammyS
Staff Emeritus