I need to find partial derivatives of a function $z = 2z + xy^2$. Usually, I'd just simplify it to $z = -xy^2$ and find that $\cfrac{\partial z}{\partial x}=-y^2$ and $\cfrac{\partial z}{\partial y}=-2yx$, however, my professor said this is not the way to solve this. I've been unsuccessful in searching for anything like this on the internet(probably due to the lack of my searching skills) so I'd appreciate any kind of information or help.
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Partial derivative with respect to what variable? – Git Gud Nov 18 '13 at 20:52
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6Your professor is being ridiculous. I can't think of a single pedagogical or mathematical reason why your approach would not be the preferred method. – Cameron Buie Nov 18 '13 at 20:54
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@GitGud to both, x and y. – user48512 Nov 18 '13 at 20:55
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1@CameronBuie Perhaps the OP is being asked to differentiate the function defined by the equation (implicitly). – Git Gud Nov 18 '13 at 20:56
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I guess you could use some sort of implicit differentiation. – Vladhagen Nov 18 '13 at 20:56
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@user48512 It would be useful to post the whole problem. – Git Gud Nov 18 '13 at 20:56
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@GitGud I did post the whole problem. It basically says 'Find partial derivatives of a given function'. – user48512 Nov 18 '13 at 20:59
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@user48512 It seems to me to be a case of implicit differentiation. – Git Gud Nov 18 '13 at 21:07
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If I understand the concept of implicit defferentiation, I'd get $\cfrac{dz}{dx} = 2\cfrac{dz}{dx} + y^2$ which basically doesn't change an answer I posted in the question. Or am I doing something wrong? – user48512 Nov 18 '13 at 21:16
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@user48512 That is correct; you're going to get the same answers you had originally. Also, as a side note, \partial generates $\partial$ in TeX, so use that to denote partial derivatives from now on. :) – Christopher Toni Nov 18 '13 at 21:21
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@ChristopherToni Thanks for the tip! – user48512 Nov 18 '13 at 21:23
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@Brett Frankel I did not know that. I wont do it again if not really needed. Thanks for this comment. – user52045 Nov 18 '13 at 21:43
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Let me make explicit why I say that there is no (substantive) reason for your professor's objection to your method. In particular, it is because (partial) differentiation is a linear operator, and in particular respects addition and scalar multiplication. I will address partial differentiation with respect to $y$ only, for brevity.
Since (partial) differentiation respects addition and scalar multiplication, then the following are equivalent: $$\frac\partial{\partial y}[z]=\frac\partial{\partial y}[2z+xy^2]\\\frac\partial{\partial y}[z]=2\frac\partial{\partial y}[z]+\frac\partial{\partial y}[xy^2]\\\frac\partial{\partial y}[z]+-2\frac\partial{\partial y}[z]=2\frac\partial{\partial y}[z]+-2\frac\partial{\partial y}[z]+\frac\partial{\partial y}[xy^2]\\-\frac\partial{\partial y}[z]=\frac\partial{\partial y}[xy^2]\\\frac\partial{\partial y}[z]=-\frac\partial{\partial y}[xy^2]\\\frac\partial{\partial y}[z]=\frac\partial{\partial y}[-xy^2]$$ That last, of course, is precisely what you'd have found if you started by the quick rearrangement of the original equation to the form $z=-xy^2,$ and then done the partial differentiation. But this is hardly surprising, as $z=2z+xy^2$ and $z=-xy^2$ are also equivalent and so applying partial differentiation to both sides of the two equations should garner another pair of equivalent equations.
You did absolutely nothing that was incorrect. It isn't the only way to proceed (in fact, there are uncountably-infinitely-many ways we could do this problem), but yours is (I contend) the best way to proceed of all of them.
Cameron Buie
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