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Lightning55 · Sep 26, 2009, 04:51 PM

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?

galactus · Sep 27, 2009, 04:26 AM

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

Unknown008 · Sep 27, 2009, 08:10 AM

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 :)

galactus · Sep 27, 2009, 08:55 AM

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.

Unknown008 · Sep 27, 2009, 09:11 AM

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

Originally Posted by galactus

The first derivative should be

Lightning55 · Sep 27, 2009, 09:14 AM

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.