Okay, I have the first derivative, but I'm not sure.
This is the original equation

y^2+2y=2x+1

and so...

dy/dx = 1/(y+1)

What I don't get is that we have to solve the first derivative at (3,7). Do I just plug in the 7 into dy/dx or did I do something wrong in getting the derivative?

And for the second derivative, I just got

d^2y/dx^2 = -1/(y+1)

Am I doing something wrong or is this right?

The first derivative should be y'=\frac{1}{y+1}. No negative.

Now, to find the second derivative, differentiate 1/(y+1):

Quotient rule:

y''=\frac{(y+1)(0)-(1)y'}{(y+1)^{2}}=\frac{-y'}{(y+1)^{2}}

But, remember that y'=1/(y+1). Sub that in:

y''=\frac{\frac{-1}{y+1}}{(y+1)^{2}}=\frac{-1}{(y+1)^{3}}

Also, from what I can tell, the point (3,7) does not lie on the curve, but (7,3) does.

I think you may have them reversed.

Therefore, if y=3, then for the first derivative we have 1/(1+3)=1/4

Good catch for the (3,7) coordinate Galactus. However, I didn't see a negative sign in front of the first derivative... :)

I was answering the question when the second derivative got me confused. Thank you, now I know how to cope with such a derivation :)

However, I didn't see a negative sign in front of the first derivative... :)

There is none in the first derivative, but there is in the second. Look close at the quotient rule.

Ah, the way you put it... it sounded as if the first derivative was good except for the negative sign..


The first derivative should be

Oh I get it. Somehow when I was using the quotient rule, I got -y instead of -y'

And no, I did not have a negative function for the first derivative.

And yes, I may have the points mixed up.

Thanks.