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- anonymous

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- schrodinger

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- TuringTest

you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?

- anonymous

if you show me just one of those partial derivatives i'll know how to find the other

- TuringTest

do I understand the problem correctly?

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## More answers

- TuringTest

I don't think I do since it seems to make no sense to me...

- 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

ooooh I see, but you are not given any more info besides
x=f(xy) ?

- TuringTest

z=f(xy)

- anonymous

yeah. that's it.

- TuringTest

well then I suppose all there is to do is apply the chain rule...

- 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

do i let xy =z then use the chain rule?

- TuringTest

okay I am confused again,\[z=f(xy)\]as in "z equals a function of x times y" correct?

- 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

yeah

- TuringTest

oh wait I invented a z, sorry

- anonymous

okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...

- TuringTest

okay let me rethink this for a sec...

- TuringTest

oh I see

- 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

would it be f ' (x) f(y) ?

- 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

that's correct.

- anonymous

correct as in they're not partial derivatives for my choices of answers

- TuringTest

so I guess then they want\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)\]

- TuringTest

this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever

- anonymous

okay thanks!

- TuringTest

welcome!

- 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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