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TuringTest Group TitleBest ResponseYou've already chosen the best response.0
you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
if you show me just one of those partial derivatives i'll know how to find the other
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
do I understand the problem correctly?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I don't think I do since it seems to make no sense to me...
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
ooooh I see, but you are not given any more info besides x=f(xy) ?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
z=f(xy)
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
yeah. that's it.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
well then I suppose all there is to do is apply the chain rule...
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
do i let xy =z then use the chain rule?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
okay I am confused again,\[z=f(xy)\]as in "z equals a function of x times y" correct?
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
actually no, i know how to do these when there are actual numbers but i'm not quite sure with the f(xy)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
oh wait I invented a z, sorry
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
okay let me rethink this for a sec...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
oh I see
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
would it be f ' (x) f(y) ?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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)???
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
that's correct.
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
correct as in they're not partial derivatives for my choices of answers
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so I guess then they want\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever
 one year ago

malcolm11235 Group TitleBest ResponseYou've already chosen the best response.0
okay thanks!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
welcome!
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago
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