If g(x,y) = ln (ye^xy) g sub x = ? My answers show it is simply y but I do not understand why. I thought since the derv of a ln is 1/(ye^xy) and since y is constant in this case the answer should be 1/e^x?????????
Note that $\displaystyle g\!\left(x,y\right)=\ln\!\left(ye^{xy}\right)=\ln\ !\left(y\right) + \ln\!\left(e^{xy}\right)=\ln\!\left(y\right)+xy$
Now, when you partially differentiate with respect to x, y is treated as a constant.
Thus, $\displaystyle \frac{\partial g}{\partial x}=\frac{\partial}{\partial x}\left[\ln y\right]+y\frac{\partial}{\partial x}\left[x\right]=0+y=\color{red}\boxed{y}$
Does this make sense?