# Is my answer for derivative right

• Apr 27th 2009, 01:35 PM
janedoe
Is my answer for derivative right
Find derivative of y= (x)/(x+y)

I used the quotient rule and then got y' on one side. My final answer is:
[(x+y)² - y] / (-2)

Is that correct?
Thanks!
• Apr 27th 2009, 01:45 PM
derfleurer
$\displaystyle y = \frac {x}{x + y}$

$\displaystyle y(x + y) - x = 0$

$\displaystyle xy + y^2 - x = 0$

$\displaystyle \frac {dy}{dx} = \frac {1 - y}{x + 2y}$
• Apr 27th 2009, 02:13 PM
janedoe
Quote:

Originally Posted by derfleurer
$\displaystyle y = \frac {x}{x + y}$

$\displaystyle y(x + y) - x = 0$

$\displaystyle xy + y^2 - x = 0$

$\displaystyle \frac {dy}{dx} = \frac {1 - y}{x + 2y}$

Now I'm confused because if you do it this way you get one answer and if you leave it as a quotient, I get a different answer.... y/[x^2 + 2xy +y^2 +x]

I can't figure out why.
• Apr 27th 2009, 02:25 PM
Jameson
The answer that he/she gave looks correct to me. Please show your work for using the quotient rule and I'll try to spot an error if there is one.
• Apr 27th 2009, 02:31 PM
janedoe
Quote:

Originally Posted by Jameson
The answer that he/she gave looks correct to me. Please show your work for using the quotient rule and I'll try to spot an error if there is one.

Ok here's one of the ways I did it:

y' = (x+y)(1) - (x)(1+y') / [(x+y)^2]
= x+y - x - xy' / [(x+y)^2]
=y-xy' / [(x+y)^2]
y'(x+y)^2 = y-xy'
y'(x+y)^2 +xy' = y
y'[(x+y)^2 +y = y
y' = y / [(x+y)^2 +x)
y' = y / [x^2 + 2xy +y^2 +x]
• Apr 27th 2009, 02:45 PM
Jameson
Your work looks fine to me. I have a feeling that both answers are equivalent, but the algebra to change your form to the other is kind of messy. Maybe someone else can show that they are equal.
• Apr 27th 2009, 03:04 PM
derfleurer
Looks right. Same thing I got using quotient rule. Wouldn't dare delve into the math to simplify them.