one more im not sure about e^(x+y) +x-y^2=3 sorry about the confusion it is e^(x+y)+x-y^2=3
Last edited by gabet16941; Jun 30th 2008 at 07:32 AM. Reason: i did not use ()
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Originally Posted by gabet16941 one more im not sure about e^x+y +x-y^2 $\displaystyle e^x+y'-2y\cdot{y'}=0$ I assume
Originally Posted by gabet16941 one more im not sure about e^x+y +x-y^2 Originally Posted by Mathstud28 $\displaystyle e^x+y'-2y\cdot{y'}=0$ I assume err.. herm.. $\displaystyle \frac{d}{dx}\left( e^x+y+x-y^2\right)=e^x+\frac{dy}{dx}+1-2y\frac{dy}{dx}$ RonL
I would say $\displaystyle e^{x+y}+(x-y)^2$ Te derivative is $\displaystyle \underbrace{(1+y')e^{x+y}}_{\text{chain rule}}+\underbrace{2(1-y')(x-y)}_{\text{chain rule}}$
Last edited by CaptainBlack; Jun 30th 2008 at 01:53 AM.
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: $\displaystyle e^x + y + x + y^2$ if you intend anything else you need to use brackets. RonL
Originally Posted by Moo I would say $\displaystyle e^{x+y}+(x-y)^2$ Te derivative is $\displaystyle \underbrace{(1+y')e^{x+y}}_{\text{chain rule}}+\underbrace{2(1-y')(x-y)}_{\text{chain rule}}$ But you are only guessing the OPs intention, that is not what they wrote?! RonL
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
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