# Partial derivatives question $w= e^{xy}\log(x^2+y^2)$

$$w= e^{xy}\log(x^2+y^2)$$

The problem is finding partial derivative with respect to $x$ of above equation. Below is what I think the answer is:

$$\frac{\partial w}{\partial x} = ye^{xy} \log(x^2 +y^2)+ e^{xy} 2x\frac{1}{(x^2+y^2)\ln 10}$$

But the back of the book says:

$$\frac{\partial w}{\partial x} = ye^{xy} \log(x^2 +y^2)+ e^{xy} 2x\frac{1}{(x^2+y^2)}$$

What am I missing or is the back of the book wrong?

-
$\log$ almost certainly means the natural logarithm, not the logarithm to base 10 –  mrf Feb 10 '13 at 20:46
I think your book uses log as ln, I have seen some books who does. –  ciceksiz kakarot Feb 10 '13 at 20:47
Indeed, in general mathematical settings, $\log$ will signify the natural logarithm. –  Pedro Tamaroff Feb 10 '13 at 20:52
oic log base e makes much more sense...I take it it's allways this way when log is writtin by itself? –  itb Feb 10 '13 at 21:02
You probably have an engineering or physics book, dont you? They tend to use $\log$ as the natural log. –  CogitoErgoCogitoSum Feb 10 '13 at 21:15

$$\frac{\partial w}{\partial x}=\frac{\partial e^{xy}}{\partial x} \cdot \ln(x^2+y^2)+e^{xy}\cdot \frac{\partial \ln(x^2+y^2)}{\partial x}=ye^{xy}\cdot \ln(x^2+y^2)+\frac{2xe^{xy}}{x^2+y^2}.$$
When referring to the decimal logarithm it's denoted with $\mathrm{lg}$. I've seen in many math books $\log$ as $\ln$.
The symbol $\mathrm{lg}$ is non-standard, though definitely use by a number of authors, so if you're going to use it, specify what you mean. Usually, people will just use $\log_{10}$. –  Avi Steiner Feb 10 '13 at 22:01