Problem in understanding chain rule.

Question

The chain rule for total derivative

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Assume that $g : \mathbb R^n \longrightarrow \mathbb R^m$ is differentiable at $a \in \mathbb R^n$, with total derivative $Df(a)$ and let $b = g(a)$ and assume that $f : \mathbb R^m \longrightarrow \mathbb R^p$ is differentiable at $b \in \mathbb R^m$, with total derivative $Df(b)$. Then the composition function $h = f \circ g : \mathbb R^n \longrightarrow \mathbb R^p$ is differentiable at $a \in \mathbb R^n$, and the total derivative $Dh(a)$ is given by

$Dh(a) = Df(b) \circ Dg(a)$, the composition of the linear functions $Df(b)$ and $Dg(a)$.

But I can't relate this concept to the chain rule involving partial derivatives as a linear combination.Please help me.

Thank you in advance.

Answer 1

Let $x,y$ be differentiable functions of $t$ and $f$ a differentiable function of $x,y$. Then, by the definition of a total derivative,

$$\begin{align}\frac d{dt}f(x(t),y(y))&=\vphantom{\cfrac11}\lim_{h\to0}\frac{f(x(t+h),y(t+h))-f(x(t),y(t))}h\\&=\vphantom{\cfrac11}\lim_{h\to0}\frac{\color{#44bb88}{f(x(t+h),y(t+h))-f(x(t),y(t+h))}+\color{#4488dd}{f(x(t),y(t+h)-f(x(t),y(t))}}h\\&=\vphantom{\cfrac11}\lim_{h\to0}\color{#44bb88}{\frac{f(x(t+h),y(t+h))-f(x(t),y(t+h))}h}+\color{#4488dd}{\frac{f(x(t),y(t+h)-f(x(t),y(t))}h}\\&=\vphantom{\cfrac11}\lim_{h\to0}\color{#44bb88}{\frac{f(x(t+h),y(t+h))-f(x(t),y(t+h))}{x(t+h)-x(t)}\frac{x(t+h)-x(t)}h}\\&+\vphantom{\cfrac11}\lim_{h\to0}\color{#4488dd}{\frac{f(x(t),y(t+h)-f(x(t),y(t))}{y(t+h)-y(t)}\frac{y(t+h)-y(t)}h}\\&=\vphantom{\cfrac11}\color{#44bb88}{\frac{\partial f(x,y)}{\partial x}\frac{dx}{dt}}+\color{#4488dd}{\frac{\partial f(x,y)}{\partial y}\frac{dy}{dt}}\end{align}$$

In general, we end up with

$\frac d{dt}f(x_1,x_2,x_3,\dots,x_k)=\sum_{p=1}^k\frac{\partial f}{\partial x_p}\frac{dx_p}{dt}$

And if we do this for one variable, we end up with

$\frac d{dt}f(x(t))=\frac{\partial f(x)}{\partial x}\frac{dx}{dt}=f'(x(t))x'(t)$

which is normal chain rule.

Answer 2

Let $f : \mathbb R^{2} \longrightarrow \mathbb R$ be a function of two real variables $x$ and $y$ which are functions of a real variable $t$.

Let us now first construct a function $g : \mathbb R \longrightarrow \mathbb R^{2}$ defined as $g(t) = (x(t),y(t))$, $t \in \mathbb R$. Let $g$ be differentiable at $a \in \mathbb R$ and $f$ be differentiable at $g(a) = (x(a),y(a))$. Now it is not hard to see that $f \circ g (t) = f(x,y)$. Then according to your post $f \circ g$ is differentiable at $a$ and therefore

$$D(g \circ f)(a) = Df(g(a)) \circ Dg(a)$$

It implies that $$D(g \circ f)(a)(1) = Df(g(a))(Dg(a)(1))$$$$ = Df(g(a))(g'(a)) = Df(g(a))(x'(a),y'(a)) = Ax'(a) + By'(a)$$ where $A = \frac {\partial f} {\partial x} {\vert}_{(x(a),y(a))}$ and $B = \frac {\partial f} {\partial y} {\vert}_{(x(a),y(a))}$. So we ultimately have

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$\frac {d}{dt}(f \circ g) {\vert}_{t=a} = \frac {d}{dt} f(x,y) {\vert}_{(x(a),y(a))} = \frac {\partial f}{\partial x} {\vert}_{(x(a),y(a))}.\frac {dx}{dt} {\vert}_{t=a} + \frac {\partial f}{\partial y} {\vert}_{(x(a),y(a))}.\frac {dy}{dt} {\vert}_{t=a}$.

Which is the required multivariable chain rule for a function of two real variables as you mentioned in your previous comment.

Answer 3

It is easiest to see with linear (well, affine) functions.

If $f(x) = Fx +a, g(x) = Gx+b$ then $h(x)=f(g(x)) = FGx + Fb +a$.

Since $Df(x) = F, Dg(x) = G$ and $Dh(x) = FG$, we see that $Dh(x) = Df(g(x)) DG(x)$.

Answer 4

What you are calling "this concept" is the chain rule, and the formula $$h'(t)=\sum_{k=1}^n {\partial f\over\partial x_k}\bigl({\bf x}(t)\bigr)\>x_k'(t)$$ for a function $$h:=f\circ{\bf x}:\quad {\mathbb R}\to{\mathbb R},\qquad t\mapsto h(t):=f\bigl({\bf x}(t)\bigr)$$ is a special case of this concept.