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?????????

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- Feb 1st 2009, 08:10 PMFrostkingpartial derivative question
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?????????

- Feb 1st 2009, 08:16 PMChris L T521
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? - Feb 1st 2009, 11:09 PMFrostkingdifferential question answer
Yes, thank you I did not realize that you should split into two natural logs and I did not know that the ln e^xy is simply xy. Thanks much! Frostking