Let $z = x^a y^b \ln(xy)$. Find $x \frac {dz} {dx} - y \frac {dz} {dy}$ in terms of $z$ I'm baffled by this question. I assume I'm meant to use the product rule to work out $\frac{dz}{dx}$ and $\frac{dz}{dy}$? 
But when I'm doing that I'm getting crazy answers that I know are wrong:
$$\frac{dz}{dx} = b y^{b-1} x^a\ln(xy)+\frac{1}{xy}xx^ay^b$$
$$\frac{dz}{dy} = b y^{2b-2}yx^ay\ln(xy)+y^{2b}x^a$$
I'm not sure why, or where I'm going wrong, but I know that this is wrong. Please help.
Thanks!
Note, the d's are 'curly' d's for the partial derivative but I wasnt sure how to type them in
 A: Your derivatives seem to be not correct. Applying the chain rule to the product of three terms, if $$z = x^a y^b \log(xy)$$ $$\frac{dz}{dx} =x^{a-1} y^b+a x^{a-1} y^b \log (x y)=x^{a-1} y^b (a \log (x y)+1)$$ $$\frac{dz}{dy}=x^a y^{b-1}+b x^a y^{b-1} \log (x y)=x^a y^{b-1} (b \log (x y)+1)$$
I am sure that you can take from here.
A: Let us assume, that $a$, $b$ are such, that the following calculations make sense.
Let us observe, that
$$
z = x^a y^b (\ln x +\ln y).
$$
Then
$$
\def\cz#1#2{\frac{\partial #1}{\partial #2}}
\cz zx=ax^{a-1}y^b(\ln x +\ln y) +x^{a-1}y^b,\quad
\cz zy=bx^{a}y^{b-1}(\ln x +\ln y)+x^{a}y^{b-1},
$$
hence
$$
x\cz zx-y\cz zy=(a-b)z.
$$
A: $z=x^ay^b\ln(xy)$. 
So, $\begin{align}\dfrac{\partial z}{\partial x}=ax^{a-1}\times y^b\times\ln(xy)+(xy)^{ab}\times\dfrac{1}{xy}\times y\end{align}\\\therefore x\dfrac{\partial z}{\partial x}=a\times x^ay^b\times \ln(xy)+(xy)^{ab}=(a\times z) +x^ay^b$. 
Similarly, 
$y\dfrac{\partial z}{\partial y}=b\times x^ay^b\times \ln(xy)+(xy)^{ab}=(b\times z) +x^ay^b$ 
$\therefore x\dfrac{\partial z}{\partial x}-y\dfrac{\partial z}{\partial y}=z(a-b)$
A: $x\dfrac{dz}{dx} = x\left(ax^{a-1}y^b\ln(xy) + x^ay^b\dfrac{y}{xy}\right) = az+x^ay^b$.
$y\dfrac{dz}{dy} = y\left(bx^ay^{b-1}\ln(xy) + x^ay^b\dfrac{x}{xy}\right) = bz + x^ay^b$.
Thus $x\dfrac{dz}{dx} - y\dfrac{dz}{dy} = (a-b)z$
