Thread: partial derivatives (multivariable calc)

z = f(xy)
∂z/∂x∂z/∂y
0
1
f '(x)
f '(y)
f '(xy)
f '(x)f(y)
f(x)f '(y)
xf '(xy)
yf '(xy)
xyf '(xy)
none of the above




0
1
f '(x)
f '(y)
f '(xy)
f '(x)f(y)
f(x)f '(y)
xf '(xy)
yf '(xy)
xyf '(xy)
none of the above




Any ideas??

My post is kind of confusing, all those f(x) and g(x) stuff under the z= are the multiple choice answers. There are 2 different questions, hence 2 different z =
Also the first one is in terms of x, the second in terms of y.
Thank You!!

Well the derivative of f is f', and the chain rule states that we multiply f' by the derivative of the argument. So for the first one, we multiply by y and the second, by x giving

a) yf '(xy)

b) xf '(xy)