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Student Support Forum: 'Partial Derivitives...' topicStudent Support Forum > General > Archives > "Partial Derivitives..."

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Author Comment/Response
Aaron Honecker
11/13/98 10:21am

>
> Apparently I didn't get my
> question across.
>
> OK. I know this:
>
> \!\(\(
D[{x*z + y*z\^2 - x}, {x, 1}]\)\)
>
> Yields this:
>
> {-1+z}
>
> Which is:
>
> \!\(\[PartialD]\/\[PartialD]x\)
>
> Which is nice, unfortunatly that is NOT what I want!
>
> I want \!\(\[PartialD]z\/\[PartialD]x\)
>
> Which I know to be:
>
> \!\(\(\(-1\) + z\)\/\(x + 2\ y\ z\)\)
>
> Because I just did it by hand.
>
> So now I'm confused, can I use this
> expensive software to do my work or
> is this #2 pencil I'm holding a better
> choice??
>
> Regards,
> ->C

Are you trying to perform partial differentiation or
implicit differentiation.
If you are performing partial differentiaion then
D[{x*z + y*z^2 - x}, {x, 1}] will perform the partial
derivative of the function with respect to x.
If you want to perform implicit differentiation with
z as a function of x and y, then
-D[{x*z + y*z^2 - x},x]/D[{x*z + y*z^2 - x},z] will
perform implicit differentiation.
The other option is to explicitly use z as a function
of x and y.
D[{x*z[x,y] + y*z[x,y]^2 - x},x] and solve for z'[x,y].

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Subject (listing for 'Partial Derivitives...')
Author Date Posted
Partial Derivitives... Christopher ... 10/22/98 09:18am
Re: Partial Derivitives... Aaron Honecker 11/13/98 10:21am
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