Say I have the following: $$f(x,a,b)=0\qquad\text{(1)}$$ Applying the implicit function theorem I get something like this: $$\frac{\partial x(a,b)}{\partial a}=\frac{f'_a}{-f'_x}\qquad\text{(2)}$$ Where $f'_a=\frac{\partial f(\cdot)}{\partial a}$ and $f'_x=\frac{\partial f(\cdot)}{\partial x}$. I am now interested in knowing how a variation in $b$ affects the effect of a variation in $a$ on $x$ (i.e., $\frac{\partial\frac{\partial x}{\partial a}}{\partial b}$).
Can I simply take the derivative of expression (2) or should I apply the implicit function theorem twice to expression (1)?
In the case where I could not, why would it be incorrect?
Many thanks!
After some research, the two methods seem equivalent in the case I consider.
Method 1: applying IFT twice
We start with $$f(x(a,b),a,b)=0\qquad \text{(1)}$$ We differentiate (1) with respect to $a$: $$f_xx_a+f_a=0\qquad \text{(2)}$$ Where $f_x$ and $f_a$ are the partial derivatives of $f(\cdot)$ with respect to $x$ and $a$ respectively. Similarly, $x_a$ is the partial derivative of $x(a,b)$ with respect to $a$. Now we differentiate (2) with respect to $b$: $$f_xx_{ab}+f_{xx}x_ax_b+f_{xa}x_b+f_{xb}x_a+f_{ab}=0\qquad\text{(3)}$$ Rearranging (3) gives: $$x_{ab}=\frac{f_{xx}x_ax_b+f_{xa}x_b+f_{xb}x_a+f_{ab}}{-f_x}\qquad\text{(4)}$$
Method 2: directly differentiating $\pmb{x_a}$
Let's start with: $$x_a=\frac{f_a}{-f_x}\qquad\text{(1')}$$ Differentiating (1') this expression with respect to $b$ gives: $$x_{ab}=\frac{-(f_{ab}+f_{xa}x_b)f_x+f_a(f_{xx}x_b+f_{xb})}{(f_x)^2}=\frac{f_{ab}+f_{xa}x_b}{-f_x}-\frac{f_a}{-f_x}\frac{f_{xx}x_b+f_{xb}}{f_x}\quad\text{(2')}$$ Injecting (1') in (2') and rearranging, we fall back on our feet: $$x_{ab}=\frac{f_{xx}x_ax_b+f_{xa}x_b+f_{xb}x_a+f_{ab}}{-f_x}$$