# diff implicit functn

• Dec 5th 2006, 09:02 AM
ashura
diff implicit functn
Find $\displaystyle \frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$,

$\displaystyle = e^{-y} \cdot - \frac {\delta y}{\delta x}$

$\displaystyle = -e^{-y} \cdot \frac {\delta y}{\delta x}$, is this correct?
• Dec 5th 2006, 09:04 AM
ThePerfectHacker
Quote:

Originally Posted by ashura
Find $\displaystyle \frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$,

$\displaystyle = e^{-y} \cdot - \frac {\delta y}{\delta x}$

$\displaystyle = -e^{-y} \cdot \frac {\delta y}{\delta x}$, is this correct?

Looks correct.
(As long as $\displaystyle y$ is a differenciable function :) )

This is my 36:):)th Post!!!
• Dec 5th 2006, 09:27 AM
CaptainBlack
Quote:

Originally Posted by ashura
Find $\displaystyle \frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$,

$\displaystyle = e^{-y} \cdot - \frac {\delta y}{\delta x}$

$\displaystyle = -e^{-y} \cdot \frac {\delta y}{\delta x}$, is this correct?

I don't know what conventions you have been taught, but to me your
notation doesn't mean anything.

It looks as though you are asked to find:

$\displaystyle \frac{d}{dx} e^{-y}$,

for which:

$\displaystyle \frac{d}{dx} e^{-y} = \frac{dy}{dx}\left(\frac{d}{dy}e^{-y}\right)=-e^{-y}\, \frac{dy}{dx}$.

RonL
• Dec 5th 2006, 09:31 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I don't know what conventions you have been taught, but to me your
notation doesn't mean anything.

It looks as though you are asked to find:

$\displaystyle \frac{d}{dx} e^{-y}$,

for which:

$\displaystyle \frac{d}{dx} e^{-y} = \frac{dy}{dx}\left(\frac{d}{dy}e^{-y}\right)=-e^{-y}\, \frac{dy}{dx}$.

RonL

Given the nature of the derivative it's probably not the same, but in Quantum we use that symbol for a functional derivative.

-Dan
• Dec 5th 2006, 11:12 AM
ashura
Thanks for the correction Ronl.