# Relationship between partial derivatives when you replace the variable you are using to derivate

Say I have a cummulative distribution function of a random variable that depends on two variables $$a_i \ \& \ a_j$$:

$$H({a}_i - {a}_j)$$

I write it's partial derivative as $$h_{a_i}({a}_i - {a}_j) = \frac{\partial H({a}_i - {a}_j)}{\partial a_i}$$

Now, if there's a formula for $$a_i$$ that depends on another variable $$b_i$$ and a number $$\delta \in (-1;1)$$ so that: $$a_i = \frac{b_i}{1- \delta^2}$$

then I can write

$$H( \frac{b_i}{1-\delta^2} - a_j)$$

I can now write a partial derivate $$h_{b_i}(\frac{b_i}{1- \delta^2} - {a}_j) = \frac{\partial H( \frac{b_i}{1-\delta^2} - a_j)}{\partial b_i}$$

What would be the relationship between $$h_{a_i}(0)$$ and $$h_{b_i}(0)$$? (The inside of the parenthesis is zero because of something outside of the scope of the question, I just need to be able to compare the partial derivatives under the same circumstances). Does the chain rule apply here so that $$h_{b_i} = h_{a_i} \cdot \frac{\partial a_i}{\partial b_i}$$?

Yes, assuming that $$a_i=\dfrac{b_i}{1-\delta^2}$$ for all $$i$$, the chain rules gives

\begin{align}\dfrac{\partial H((b_i-b_j)/(1-\delta^2))}{\partial b_i}&=\dfrac{\partial[(b_i-b_j)/(1-\delta^2)]}{\partial b_i}\cdot\left.\dfrac{\partial H(a_i-a_j)}{\partial a_i}\right\rvert_{a_i:=b_i/(1-\delta^2)\\a_j:=b_j/(1-\delta^2)}\\[2ex]&=\dfrac{1}{1-\delta^2}\cdot\left.\dfrac{\partial H(a_i-a_j)}{\partial a_i}\right\vert_{a_i:=b_i/(1-\delta^2)\\a_j:=b_j/(1-\delta^2)}\end{align}

PS: I would avoid using the designations $$h_{a_i}$$ or $$h_{b_i}$$ as they are a source of confusion, since $$H$$ is actually a monovariate function which is being composed with a bivariate argument $$(a_i-a_j)$$.

At the very least, designate $$G(a_i,a_j):=H(a_i-a_j)$$ and $$F(b_i,b_j,\delta):=H((b_i-b_j)/(1-\delta^2))$$ so you can have expression for the partial derivatives with respect to the arguments of these multivariate functions. $$F^{(1,0,0)}(b_i,b_j,\delta)= \dfrac{G^{(1,0)}(b_i/(1-\delta^2),b_j/(1-\delta_2))}{1-\delta^2}$$

PPS: The chain rules for the partial derivative with respect to $$\delta$$ is:

$$\small F^{(0,0,1)}(b_i,b_j,\delta) = \dfrac{2\delta}{(1-\delta^2)^2}\left[b_i~G^{(1,0)}\big(b_i/(1-\delta^2), b_j/(1-\delta^2)\big)-b_j~G^{(0,1)}\big(b_i/(1-\delta^2), b_j/(1-\delta^2)\big)\right]$$