diff implicit functn

**ashura** · December 5th 2006, 09:02 AM

Find $\frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$ ,

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

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

**ThePerfectHacker** · December 5th 2006, 09:04 AM

Originally Posted by ashura

Find $\frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$ ,

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

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

Looks correct.
(As long as $y$ is a differenciable function

)

This is my 36

th Post!!!

**CaptainBlack** · December 5th 2006, 09:27 AM

Originally Posted by ashura

Find $\frac {\delta y}{\delta x} \, \left ( e^{-y} \right )$ ,

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

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

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

for which:

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

RonL

**topsquark** · December 5th 2006, 09:31 AM

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:

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

for which:

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

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

**ashura** · December 5th 2006, 11:12 AM

Thanks for the correction Ronl.

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