I am trying to take the derivative of this function. First, I want to take the derivative of the original function, $F(x,\eta)$, with respect to $\eta$:
$$F(x,\eta)=\frac{\sqrt{2}\pi}{3}g(1,\eta)\frac{x}{[1+C(\eta)(x-1)]^3}$$
Do not worry about what $g(1,\eta)$ and $C(\eta)$ are, just know that they are both just functions of $\eta$
So, if I want to that the derivative with respect to $\eta$, this is my attempt. I was hoping someone could check the work to make sure it looks ok.
$$\frac{\partial F}{\partial \eta}=\frac{\sqrt{2}\pi}{3}\left[g'(1,\eta)\frac{x}{[1+C(\eta)(x-1)]^3}+g(1,\eta)\left(\frac{x}{3[1+C(\eta)(x-1)]^2}*\frac{1}{C'(\eta)(x-1)}\right)\right]$$
If this looks ok, I'll try and take the derivative with respect to $x$ too.
The second term has some problems, it should be something like
$$ \frac{\partial F}{\partial \eta} = \frac{\sqrt{2}\pi}{3}\left[g'(1,\eta) \frac{x}{[1 + C(\eta)(x - 1)]^3} - 3g(1,\eta)\left(\frac{x(x -1 )C'(\eta)}{[1 + C(\eta)(x-1)]^4}\right)\right] $$
which just comes from
$$ \frac{{\rm d}}{{\rm d}\eta} \frac{1}{[f(\eta)]^3} = \frac{{\rm d}}{{\rm d}\eta} [f(\eta)]^{-3} = -3[f(\eta)]^{-4}\frac{{\rm d}f(\eta)}{{\rm d}\eta} = -3\frac{1}{[f(\eta)]^4} \frac{{\rm d}f(\eta)}{{\rm d}\eta} $$
Similarly, if you take the derivative w.r.t. $x$ you should obtain
$$ \frac{\partial F}{\partial x} = \frac{\sqrt{2}\pi}{3}g(1,\eta)\left[\frac{1}{[1 + C(\eta)(x - 1)]^3} -\frac{3 x C(\eta)}{[1 + C(\eta)(x -1)]^4}\right] $$