# IMolicit differentiation

• Jun 29th 2008, 07:48 PM
gabet16941
IMolicit differentiation
one more im not sure about

e^(x+y) +x-y^2=3

it is e^(x+y)+x-y^2=3
• Jun 29th 2008, 08:17 PM
Mathstud28
Quote:

Originally Posted by gabet16941
one more im not sure about

e^x+y +x-y^2

$e^x+y'-2y\cdot{y'}=0$

I assume
• Jun 29th 2008, 11:13 PM
CaptainBlack
Quote:

Originally Posted by gabet16941
one more im not sure about

e^x+y +x-y^2

Quote:

Originally Posted by Mathstud28
$e^x+y'-2y\cdot{y'}=0$

I assume

err.. herm..

$\frac{d}{dx}\left( e^x+y+x-y^2\right)=e^x+\frac{dy}{dx}+1-2y\frac{dy}{dx}$

RonL
• Jun 30th 2008, 12:15 AM
Moo
I would say $e^{x+y}+(x-y)^2$

Te derivative is $\underbrace{(1+y')e^{x+y}}_{\text{chain rule}}+\underbrace{2(1-y')(x-y)}_{\text{chain rule}}$
• Jun 30th 2008, 01:50 AM
CaptainBlack
Quote:

Originally Posted by gabet16941
one more im not sure about

e^x+y +x-y^2

Please use brackets to make the expression unambiguous. The normal precedence of operators means that without brackets this represents:

$e^x + y + x + y^2$

if you intend anything else you need to use brackets.

RonL
• Jun 30th 2008, 01:53 AM
CaptainBlack
Quote:

Originally Posted by Moo
I would say $e^{x+y}+(x-y)^2$

Te derivative is $\underbrace{(1+y')e^{x+y}}_{\text{chain rule}}+\underbrace{2(1-y')(x-y)}_{\text{chain rule}}$

But you are only guessing the OPs intention, that is not what they wrote?!
RonL
• Jun 30th 2008, 09:16 AM
Moo
Quote:

Originally Posted by CaptainBlack
But you are only guessing the OPs intention, that is not what they wrote?!
RonL

Isn't "would" a sign of "condition" ? lol

It doesn't matter actually ^^

See ya