anonymous
  • anonymous
p
Mathematics
schrodinger
  • schrodinger
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TuringTest
  • TuringTest
you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?
anonymous
  • anonymous
if you show me just one of those partial derivatives i'll know how to find the other
TuringTest
  • TuringTest
do I understand the problem correctly?

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TuringTest
  • TuringTest
I don't think I do since it seems to make no sense to me...
anonymous
  • anonymous
the partial derivative of z with repsect to x. del z/ del x or die z / die x . that symbol that is used for partial derivatives that looks like a two kind of
TuringTest
  • TuringTest
ooooh I see, but you are not given any more info besides x=f(xy) ?
TuringTest
  • TuringTest
z=f(xy)
anonymous
  • anonymous
yeah. that's it.
TuringTest
  • TuringTest
well then I suppose all there is to do is apply the chain rule...
anonymous
  • anonymous
so i can start off by going z - f(xy) = 0 then i need to find the partial derivative of x, so f_x then f_y and f_z and del z/ del x = -f_x / f_z
anonymous
  • anonymous
do i let xy =z then use the chain rule?
TuringTest
  • TuringTest
okay I am confused again,\[z=f(xy)\]as in "z equals a function of x times y" correct?
anonymous
  • anonymous
actually no, i know how to do these when there are actual numbers but i'm not quite sure with the f(xy)
anonymous
  • anonymous
yeah
TuringTest
  • TuringTest
oh wait I invented a z, sorry
anonymous
  • anonymous
okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...
TuringTest
  • TuringTest
okay let me rethink this for a sec...
TuringTest
  • TuringTest
oh I see
anonymous
  • anonymous
i know what i want to do but i'm not quite sure how to do it. the only thing throwing me off is f(xy) so if i take the partial derivative of that with respect to x, i hold y constant. if it was a normal expression no problem but since its not i'm a little confused
anonymous
  • anonymous
would it be f ' (x) f(y) ?
TuringTest
  • TuringTest
my reasoning gives\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}f(xy)=\frac{\partial f}{\partial x}\cdot\frac{\partial }{\partial x}(xy)=y\frac{\partial f}{\partial x}=yf_x(xy)\]but the choices in your answer are for f' (whole derivatives on f, not partial)???
anonymous
  • anonymous
that's correct.
anonymous
  • anonymous
correct as in they're not partial derivatives for my choices of answers
TuringTest
  • TuringTest
so I guess then they want\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)\]
TuringTest
  • TuringTest
this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever
anonymous
  • anonymous
okay thanks!
TuringTest
  • TuringTest
welcome!
anonymous
  • anonymous
if the question's asking for partial derivatives then z = f(x*y) makes sense. If the question's asking for the total differential wrt x or y than they could just be asking about the chain rule and z= f(x, y)

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