Ok the chain rule for single variable calculus is as follow: if $ f:I \to \mathcal{R}$ is differentiable at $d \in I$, $f(I) \subset J $and $g: J \to \mathcal{R}$ differentiable at $f(d)$. then $g\circ f(x)$ is differentiable at $d$ and the derivative is $g'(f(d))f'(d)$.
I memorized the proof that uses the caratheodory theorem. But I really still don't understand why we need to go such a trouble to prove this theorem.
I really don't understand what is this $dx,dy$ and those kind of stuff. like for example, why can you treat them like numbers and do $dy= y'(x)dx$ and then substitute that into an integral. I thought $\frac{dy}{dx}$as whole is a symbole.
I do know the epsilon delta definition of continuity, differentiability.
This just limit my understanding in multivariable calculus very much as well.
Like when they say that $\frac{df(x(t),y(t))}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}$. I don't understand it even though i know how to compute it.
DO you see my problem? Can you please enlighten me?
Best Answer
I suspect you are just getting lost in the collection of overused symbols $x,y,f$.
The chain rule for $g \circ f$ is $D (g \circ f) (x) = Dg(f(x))Df(x)$.
Suppose $g(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$, then $Dg(t) = \begin{bmatrix} {\partial x(t) \over \partial t} \\ {\partial y(t) \over \partial t} \end{bmatrix}$ and you have $Df(z) = \begin{bmatrix} {\partial f(z) \over \partial z_1} &{\partial f(z) \over \partial z_2} \end{bmatrix}$
so $D (f \circ g) (t) = Df(g(t)) Dg(t) = \begin{bmatrix} {\partial f(\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}) \over \partial z_1} & {\partial f(\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}) \over \partial z_2} \end{bmatrix} \begin{bmatrix} {\partial x(t) \over \partial t} \\ {\partial y(t) \over \partial t} \end{bmatrix} = {\partial f(\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}) \over \partial z_1} {\partial x(t) \over \partial t} + {\partial f(\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}) \over \partial z_2} {\partial y(t) \over \partial t} $