Chain rules and calculus

Question

I've forgotten simple chain rules of calculus. Let $G(x,y) = x - \frac{F(x,y)}{1+F(x,y)}$ where $F(x,y) = h(x,y) \frac{x}{1-x}$. $x$ and $y$ are independent. What is the second derivative of $G$ wrt $h$?

Answer 1

$\frac{\partial F}{\partial h}=\frac{\partial F}{\partial x}\frac{\partial x}{\partial h}+\frac{\partial F}{\partial y}\frac{\partial y}{\partial h}$ this is the basic rule. we have two facts:- the first one that x is independent, $\frac{\partial x}{\partial h}=0$ the second one that y is independent, $\frac{\partial y}{\partial h}=0$ hence$\frac{\partial F}{\partial h}=0$. Therefore, $\frac{\partial G}{\partial h}=0$, if you differentiated one more time you will get zero again.

Answer 2

Let $p=h(x,y)$ and $q=F(x,y).$ Then we have that $$q=\frac{px}{1-x}.$$ Thus, we have that $$r=G(x,y)=x-\frac{q}{1+q}=x-\frac{\frac{px}{1-x}}{1+\frac{px}{1-x}}=x-\frac{px}{1-x+px},$$ which is equal to $$x-\frac{1}{1+\frac{1-x}{px}},$$ provided $px\ne 0.$

Therefore, to find the second derivative of $r$ with respect to $p,$ we consider $x$ a function of $p$ and $y,$ with these last variables now independent. Then $r$ depends on $p,\,y.$ Hence we have $$r'_p=x'_p-\frac{\left(\frac{1-x}{px}\right)'_p}{\left(1+\frac{1-x}{px}\right)^2},$$ or $$x'_p-\frac{\frac{1}{p^2}-\frac{1}{xp^2}-\frac{x'_p}{px^2}}{\left(1+\frac{1-x}{px}\right)^2}.$$ Simplify this and go for the second derivative.

If $px=0,$ then we have that $r=x,$ and therefore we have $r'_p=x'_p$ for this case, so that $r''_p=x''_p.$