Clarification on derivative of multi-variable function.

Question

I was reading this answer to a previous question which concerns calculating the second derivative of implicit functions. I've got a problem with how he expands the expression. It goes as follows:

We compute, using the standard formula for $(f/g)'$, $(f/g)′=(f'g−fg′)/f^2$, and the fact that ${d \over dx} G(x,y(x))=G_x+G_{yy} \tag{$\star$}$ for any sufficiently differentiable function of two variables $G(x,y)$: $-\frac{d}{dx}(F_x / F_y) = -((F_{xx} + F_{xy}y')F_y - F_x(F_{yx} + F_{yy}y'))/F^2_y, \tag{4}$

My problem is ${d \over dx}F_x = F_{xx} + F_{yx}y'\tag{$\star\star$}$

I know is correct due to $(\star)$, but on the other hand, isn't it true that $F_x = {d F \over d x }$, and by defenition ${d \over dx}F_x = {d^2 F \over d x^2} = F_{xx}$, which is clearly different from $(\star \star)$. What am I missing?

Answer 1

$F$ is function of x and y.

$F_x = \frac {\partial F}{\partial x}$ is the partial derivative with respect to x

The total derivative of $F$ with respect to $x$ is $\frac {\partial F}{\partial x} + \frac {\partial F}{\partial y} \frac {dy}{dx} = F_x + F_y y'$

The total derivative of $F_x = \frac {\partial F_x}{\partial x} + \frac {\partial F_x}{\partial y} \frac {dy}{dx} =\frac {\partial^2 F}{\partial x^2} + \frac {\partial^2 F}{\partial x\partial y} \frac {dy}{dx} = F_{xx} + F_{xy} y'$