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

Question

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

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?

Answer 1

$$\frac{\partial w}{\partial x}=\frac{\partial e^{xy}}{\partial x} \cdot \ln(x^2+y^2)+e^{xy}\cdot \frac{\partial \ln(x^2+y^2)}{\partial x}=ye^{xy}\cdot \ln(x^2+y^2)+\frac{2xe^{xy}}{x^2+y^2}. $$

When referring to the decimal logarithm it's denoted with $\mathrm{lg}$. I've seen in many math books $\log$ as $\ln$.