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z=f(x/y) Using partial derivatives (Multivariable calculus) find dz/dy. I know that dz/dx is f'(x/y)/y. I am given the following choices:

A) 0

B) 1

C) f '(x/y)/x

D) f '(x/y)/y

E) xf '(x/y)/x

F) yf '(x/y)/y

G) xf '(x/y)/x^2

H) yf '(x/y)/y^2

I) xf '(x/y)/y^2

J) yf '(x/y)/x^2

K) -xf '(x/y)/x^2

L) -yf '(x/y)/y^2

M) -xf '(x/y)/y^2

N) -yf '(x/y)/x^2

O) none of the above

I know for sure that D is not it because that is dz/dx. I also assume that anything which has values that cancel out such as xf'(x/y)/x are not it, so that takes out E,F. In that case, G & H are equivalent to C & D, so all those are out of the question. Even though I don't think A and B are right, I'm not 100%, so I'll leave them in.

My options are now A B I J K L M N O. Can someone help me reason through this and figure out which one is right, if any?

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Why can't you just calculate it, instead of trying to check all those options? – Javier Sep 27 '12 at 1:03
up vote 0 down vote accepted

Let $u(x, y) = \frac{x}{y}$, then $ z =f(u(x,y))$. do you remember the chain rule?

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We had not gone over the chain rule for multiple variables yet, but I understand now. Thanks! – Jared Oct 3 '12 at 16:57

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