Question

Can anyone do the partial derivatives of (e^y)sinxy?

Answer 1

\[f_x = (e^y)\frac{ ∂ }{ ∂x }\sin(xy)+\sin(xy)\frac{ ∂ }{ ∂x }(e^y)\] \[f_y = (e^y)\frac{ ∂ }{ ∂y }\sin(xy)+\sin(xy)\frac{ ∂ }{ ∂y }(e^y)\]

Answer 2

Treating "y" as a constant for the first one, whilst treating "x" as a constant for the second partial derivative.

Answer 3

\[f_x = e^y[y \cos(xy)]+ \sin(xy)[e^y \times 0] = ye^ycos(xy)\]

Answer 4

My partial in terms of x was the same as yours so thanks for verifying that. My partial in terms of y was \[e^yxcos(xy)-ye^ysin(xy)\]