1. ## Partial Derivatives

Got another question, this time its a bit harder and im not sure how to get the partial derivatives
The function is:
f(x,y) = e^1/2xy

e to the power of one half(x times y)

Need help with finding:
fx
fy
fxx
fyy
fxy
fyx

Thanks!

2. Hello, Hockey_Guy14!

Find the first and second partial derivatives:

. . . $f(x,y) \:=\: e^{\frac{1}{2}xy}$

. . $\begin{array}{ccccc}f_x &=& e^{\frac{1}{2}xy}\cdot\frac{1}{2}y &=& \frac{1}{2}y\,e^{\frac{1}{2}xy} \\ \\[-3mm]
f_y &=& e^{\frac{1}{2}xy}\cdot\frac{1}{2}x &=& \frac{1}{2}x\,e^{\frac{1}{2}xy} \end{array}$

. . $\begin{array}{ccccc}f_{xx} &=& \frac{1}{2}y\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}y &=& \frac{1}{4}y^2\,e^{\frac{1}{2}xy} \\ \\[-3mm]
f_{yy} &=& \frac{1}{2}x\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}x &=& \frac{1}{4}x^2\,e^{\frac{1}{2}xy} \end{array}$

. . $\begin{array}{ccccc}
f_{xy} &=& \frac{1}{2}y\cdot e^{\frac{1}{2}xy}\cdot\frac{1}{2}x + \frac{1}{2}\,e^{\frac{1}{2}xy} & = &\frac{1}{4}e^{\frac{1}{2}xy}\,(xy + 2) \\ \\[-3mm]
f_{yx} &=& \frac{1}{2}x\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}y + \frac{1}{2}e^{\frac{1}{2}xy} &=& \frac{1}{4}e^{\frac{1}{2}xy}\,(xy + 2)
\end{array}$