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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.

N. F. Taussig
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Kwame Brown
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1 Answers1

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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 $a+b-cd-ce=0$

$$\frac{\delta\log\left(\frac X Y\right)}{\delta\log\left(\frac{A}{B}\right)}=\frac 1{a-cd}$$

user
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  • I have always thought that differentials and derivates are different, in that we cannot think of the derivative as a fraction. Is this not what you have done here? – Kwame Brown May 17 '23 at 17:08
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    Yes, they are indeed different but by definition the ratio between differential is the derivative. You can find a detailed discussion here https://math.stackexchange.com/q/23902/505767 – user May 17 '23 at 17:19