I have the following function $$ a\text{log}\left(X\right)+b\text{log}\left(y\right)=c\left[d\text{log}\left(X\right)+e\text{log}\left(y\right)\right]+\text{log}\left(\frac{A}{B}\right) $$ I wish to compute the following derivative: $$ \frac{\partial\log\left(\frac{X}{Y}\right)}{\partial\log\left(\frac{A}{B}\right)} $$
I have tried numerous techniques, but am not sure how to proceed. I imagine the implicit function theorem is appropriate here, but I am unable to write the LHS in terms of $\log\left(\frac{X}{Y}\right)$ given the presence of $a$ and $b$ coefficients. Any pointers on this are much appreciated.
Using $\delta$ as a symbol for differential to avoid confusion with constant $d$, we have
$$a\log\left(X\right)+b\log\left(Y\right)=c\left[d\log\left(X\right)+e\log\left(Y\right)\right]+\log\left(\frac{A}{B}\right)$$
$$a\delta\log\left(X\right)+b\delta\log\left(Y\right)=c\left[d\delta\log\left(X\right)+e\delta\log\left(Y\right)\right]+\delta\log\left(\frac{A}{B}\right)$$
$$(a-cd)\delta\log\left(X\right)+(b-ce)\delta\log\left(Y\right)=\delta\log\left(\frac{A}{B}\right)$$
$$(a-cd)\delta\log\left(\frac X Y\right)+(a+b-cd-ce)\delta\log\left(Y\right)=\delta\log\left(\frac{A}{B}\right)$$
$$(a-cd)\delta\log\left(\frac X Y\right)=\delta\log\left(\frac{A}{B}\right)-(a+b-cd-ce)\delta\log\left(Y\right)$$
$$\frac{\delta\log\left(\frac X Y\right)}{\delta\log\left(\frac{A}{B}\right)}=\frac{1-(a+b-cd-ce)\frac{\delta\log\left(Y\right)}{\delta\log\left(\frac{A}{B}\right)}}{a-cd}$$
with $a-cd \neq 0$ and for $a+b-cd-ce=0$
$$\frac{\delta\log\left(\frac X Y\right)}{\delta\log\left(\frac{A}{B}\right)}=\frac 1{a-cd}$$