Finding general solution $\frac{d^2f}{dx^2}$ for an implicit function

Question

Let's say I have a function $F:\mathbb R \times \mathbb R \rightarrow \mathbb R$, which meets the requirements of the implicit function theorem. Now I would like to find a general expression for: $$\frac{d^2f}{dx^2}$$

From the theorem I know that $$\frac{df}{dx} = - \Big(\frac{\partial F}{\partial y} \Big )^{-1} \frac{\partial F}{\partial x}$$ So I went on to say that $$\frac{d^2f}{dx^2} = - \frac{\frac{d}{dx}(\frac{\partial F}{\partial x})*\frac{\partial F}{\partial y}-\frac{\partial F}{\partial x}*\frac{d}{dx}(\frac{\partial F}{\partial y})}{(\frac{\partial F}{\partial y})^2}$$ By using the quotient rule. Is this correct, and if so: How can I simplify this any further? I found online that $$\frac{d^2f}{dx^2}=-\frac {\frac{\partial^2 F}{\partial x^2}\left(\frac{\partial F}{\partial y}\right)^2 -2·\frac{\partial^2 F}{\partial x\partial y}·\frac{\partial F}{\partial y}·\frac{\partial F}{\partial x} +\frac{\partial^2 F}{\partial y^2}\left(\frac{\partial F}{\partial x}\right)^2} {\left(\frac{\partial F}{\partial y}\right)^3}$$

which is correct I guess, but I do not quite know how to get there. Any help is greatly appreciated!

Answer 1

Write $F = F(x,y)$ so that

$$ F(x, f(x)) = 0. $$

By differentiating once with respect to $x$ and using the chain rule, we get

$$ \frac{\partial F}{\partial x}(x,f(x)) + \frac{\partial F}{\partial y}(x,f(x)) f'(x) = 0. $$

Hence,

$$ f'(x) = - \frac{\frac{\partial F}{\partial x}(x,f(x))}{\frac{\partial F}{\partial y}(x,f(x))}. $$

To differentiate it again, we need to calculate the derivatives of the numerator and the denominator and use the quotient rule. For both the numerator and the denominator, we will need to use again the chain rule. Let me demonstrate it for the numerator:

$$ \frac{d}{\partial x} \left( \frac{\partial F}{\partial x}(x,f(x)) \right) = \frac{\partial^2 F}{\partial x^2}(x,f(x)) + \frac{\partial^2 F}{\partial y \partial x}(x,f(x)) f'(x) = \\ \frac{\frac{\partial F}{\partial y}(x,f(x)) \frac{\partial^2 F}{\partial x^2}(x,f(x)) - \frac{\partial F}{\partial x}(x,f(x)) \frac{\partial^2 F}{\partial y \partial x} (x,f(x))}{\frac{\partial F}{\partial y}(x,f(x))}. $$

If you do the same for the denominator and do some arithmetic, you'll get the formula you quoted.

Answer 2

HINT:

Let $\displaystyle F_x(x,y)=\frac{\partial F(x,y)}{\partial x}$ and $\displaystyle F_y(x,y)=\frac{\partial F(x,y)}{\partial y}$.

$$\begin{align} \frac{d}{dx}F_x(x,f(x))&=F_{xx}(x,f(x))+F_{xy}(x,f(x))\frac{df(x)}{dx}\\\\ &=F_{xx}(x,f(x))-F_{xy}(x,f(x))\frac{F_x(x,f(x))}{F_y(x,f(x))} \end{align}$$

Answer 3

$F(x,f(x))=0$ and so $$F_1 + F_2 f'=0 \ \ \text{ or}\ \ f'=-\frac{F_1}{F_2}$$ by the chain rule.

Now differentiate again, using chain rule and product rule, $$F_{11}+F_{12}f'+F_{21}f'+F_{22}(f')^2+F_2f''=0$$ which we can re-arrange and substitute for $f'$ to get the formula you quote.