Student Support Forum: 'Partial Derivitives...' topicStudent Support Forum > General > "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]. URL: ,

 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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