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malcolm11235
 2 years ago
p
malcolm11235
 2 years ago
p

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0do I understand the problem correctly?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I don't think I do since it seems to make no sense to me...

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0the 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
 2 years ago
Best ResponseYou've already chosen the best response.0ooooh I see, but you are not given any more info besides x=f(xy) ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0well then I suppose all there is to do is apply the chain rule...

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0so 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

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0do i let xy =z then use the chain rule?

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0oh wait I invented a z, sorry

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0okay let me rethink this for a sec...

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0would it be f ' (x) f(y) ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0my 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)???

malcolm11235
 2 years ago
Best ResponseYou've already chosen the best response.0correct as in they're not partial derivatives for my choices of answers

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

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0if 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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