t-derivative of $f\left(\frac{x-g(t)}{h(t)}\right)$

Question

I'm certain I'm just overthinking this, I've been staring at this for a while! My question is, if a function was written in the form $f\left(\frac{x-g(t)}{h(t)}\right)$, what would the t-derivative be? I know that if the function was written as the form $a(x, g(t), h(t))$, it'd be done via the chain rule and would be $$\frac{\partial a}{\partial t} = \frac{\partial a}{\partial g}\frac{dg}{dt} + \frac{\partial a}{\partial h}\frac{dh}{dt}$$ but in my eyes now, $f\left(\frac{x-g(t)}{h(t)}\right)$ isn't dependent on $h(t)$ but rather $h^{-1}(t)$. Would this affect it? Or have I just been staring at this for far too long?

Answer 1

Check this link for "Faa' di Bruno" and "Leibniz" formulae: https://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_to_nth_order
Using the "Faa' di Bruno" and "Leibniz" formulae, you shall first assume $s(t)=\frac{x-g(t)}{h(t)}$. Now use "Faa' di Bruno" formula to evaluate $f(s(t))$. Now for evaluating this derivative using "Faa' di Bruno" formula you also need the n-th derivative to $s(t)$. For finding this, assume $r_1(t)=x-g(t)$ and $r_2(t)=\frac{1}{h(t)}$ and use "Leibniz" rule to find the n-th derivative to $s(t)=r_1(t)r_2(t)$ and then you're done!