Question

The partial derivative with respect to y for z=e^y sin(xy) is:
zy= xe^y cos(xy) + e^y sin(xy)
How do i get to this answer? Thanks.

Answer 1

d z/dy= d/dy e^y sin(xy)
= sin xy * e^y + cos(xy)*d(xy)/dy *e^y

but d(xy)/dy=x

so there it is.

Answer 2

But why wouldnt the answer be just
xe^y cos(xy)? if the partial derivative with respect to x is just ye^y cos(xy).