View Full Version : Implicit Differentiation and Second Derivative
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..
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.