# Implicit Differentiation - Check my work?

• Jan 1st 2010, 12:24 PM
pocketasian
Implicit Differentiation - Check my work?
Hello, I'm doing an AP Cal AB practice test, and I'm having trouble with one of the problems. I think that I'm solving the problem correctly, but I must be doing something wrong since my answer is not one of the choices. Seeing how this was on a previous AP exam, I don't think it's likely that the choices were written incorrectly. Will you check over my work?
http://i13.photobucket.com/albums/a2...g?t=1262380909
• Jan 1st 2010, 12:50 PM
Jester
Quote:

Originally Posted by pocketasian
Hello, I'm doing an AP Cal AB practice test, and I'm having trouble with one of the problems. I think that I'm solving the problem correctly, but I must be doing something wrong since my answer is not one of the choices. Seeing how this was on a previous AP exam, I don't think it's likely that the choices were written incorrectly. Will you check over my work?
http://i13.photobucket.com/albums/a2...g?t=1262380909

I believe your answer is tha same as (a) - factor the negative 1!
• Jan 1st 2010, 01:10 PM
e^(i*pi)
Quote:

Originally Posted by pocketasian
Hello, I'm doing an AP Cal AB practice test, and I'm having trouble with one of the problems. I think that I'm solving the problem correctly, but I must be doing something wrong since my answer is not one of the choices. Seeing how this was on a previous AP exam, I don't think it's likely that the choices were written incorrectly. Will you check over my work?
http://i13.photobucket.com/albums/a2...g?t=1262380909

It looks like your answer is equivalent to A. If you factor out -1 in your numerator you should get the same syntax as answer A (both are correct still)
• Jan 1st 2010, 02:58 PM
pocketasian
So even though there aren't parentheses around 2x+y in choice A, it can still be considered as having a -1 factored out? I would think -(2x+y) would make more sense...
• Jan 1st 2010, 03:03 PM
e^(i*pi)
Quote:

Originally Posted by pocketasian
So even though there aren't parentheses around 2x+y in choice A, it can still be considered as having a -1 factored out? I would think -(2x+y) would make more sense...

Option A is unclear. Do you mean $\displaystyle - \frac{2x+y}{x+3y^2}$ or

$\displaystyle \frac{-2x+y}{x+3y^2}$
• Jan 1st 2010, 03:13 PM
pocketasian
Quote:

Originally Posted by e^(i*pi)
Option A is unclear. Do you mean $\displaystyle - \frac{2x+y}{x+3y^2}$ or

$\displaystyle \frac{-2x+y}{x+3y^2}$

It's written as
$\displaystyle - \frac{2x+y}{x+3y^2}$
I guess there's a difference then?
• Jan 1st 2010, 04:12 PM
Defunkt
Quote:

Originally Posted by pocketasian
It's written as
$\displaystyle - \frac{2x+y}{x+3y^2}$
I guess there's a difference then?

Yes, there is.

$\displaystyle -\frac{2x+y}{x+3y^2} = (-1)\frac{2x+y}{x+3y^2} = \frac{(-1)(2x+y)}{x+ey^2} = \frac{-2x-y}{x+3y^2} \neq \frac{-2x+y}{x+3y^2}$