Can the chain rule be proven by math induction? I need to prove the chain rule for a math project and I am wondering if it can be proven by math induction. If not, how can this rule be proven?
 A: I will outline a proof here that is really adapted from an old calculus text by James Stewart. This is not from an analysis book but from a standard calculus text; thus, the rigor is still maintained but not quite as terse. A proof from Rudin is offered at the end for a different style. 

Prefatory comments: Recall that if $y=f(x)$ and $x$ changes from $a$ to $a+\Delta x$, we define the increment of $y$ as
$$
\Delta y=f(a+\Delta x)-f(a).
$$
According to the definition of a derivative, we have
$$
\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=f'(a).
$$
So if we denote by $\epsilon$ the difference between the difference quotient and the derivative, we obtain
$$
\lim_{\Delta x\to 0}\epsilon = \lim_{\Delta x\to 0}\left(\frac{\Delta y}{\Delta x}-f'(a)\right)=f'(a)-f'(a)=0.
$$
But
$$
\epsilon=\frac{\Delta y}{\Delta x}-f'(a)\quad\Longrightarrow\quad\Delta y = f'(a)\Delta x+\epsilon\Delta x.
If we define $\epsilon$ to be $0$ when $\Delta x=0$, then $\epsilon$ becomes a continuous function of $\Delta x$. Thus, for a differentiable function $f$, we can write
$$
\Delta y=f'(a)\Delta x+\epsilon\Delta x\quad\text{where}\quad\epsilon\to 0\quad\text{as}\quad\Delta x\to 0,\tag{1}
$$
and $\epsilon$ is a continuous function of $\Delta x$. This property of differentiable functions is what enables us to prove the Chain Rule.

Proof: Suppose $u=g(x)$ is differentiable at $a$ and $y=f(u)$ is differentiable at $b=g(a)$. If $\Delta x$ is an increment in $x$ and $\Delta u$ and $\Delta y$ are the corresponding increments in $u$ and $y$, then we can use equation $(1)$ to write
$$
\Delta u=g'(a)\Delta x+\epsilon_1\Delta x=[g'(a)+\epsilon_1]\Delta x,\tag{2}
$$
where $\epsilon_1\to 0$ as $\Delta x\to 0$. Similarly,
$$
\Delta y = f'(b)\Delta u+\epsilon_2\Delta u=[f'(b)+\epsilon_2]\Delta u,\tag{3}
$$
where $\epsilon_2\to 0$ as $\Delta u\to 0$. If we now substitute the expression for $\Delta u$ from equation $(2)$ into equation $(3)$, we get
$$
\Delta y=[f'(b)+\epsilon_2][g'(a)+\epsilon_1]\Delta x
$$
so
$$
\frac{\Delta y}{\Delta x}=[f'(b)+\epsilon_2][g'(a)+\epsilon_1].
$$
As $\Delta x\to 0$, equation $(2)$ shows that $\Delta u\to 0$. So both $\epsilon_1\to0$ and $\epsilon_2\to0$ as $\Delta x\to 0$. Therefore,
\begin{align}
\frac{dy}{dx} &= \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\[1em]
              &= \lim_{\Delta x\to 0}[f'(b)+\epsilon_2][g'(a)+\epsilon_1]\\[1em]
              &= f'(b)g'(a)=f'(g(a))g'(a).
\end{align}
This proves the Chain Rule (see below for a much more terse treatment). $\Box$

Proof of Chain Rule in Baby Rudin: The Chain Rule is first stated as a theorem:
Theorem. Suppose $f$ is continuous on $[a,b], f'(x)$ exists at some point $x\in[a,b], g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$. If
$$
h(t) = g(f(t))\qquad(a\leq t\leq b),
$$
then $h$ is differentiable at $x$, and 
$$
h'(x) = g'(f(x))f'(x).\tag{1}
$$
Proof. Let $y=f(x)$. By the definition of the derivative, we have
$$
f(t)-f(x)=(t-x)[f'(x)+u(t)],\tag{2}
$$
and
$$
g(s)-g(y)=(s-y)[g'(y)+v(s)],\tag{3}
$$
where $t\in[a,b], s\in I$, and $u(t)\to 0$ as $t\to x, v(s)\to 0$ as $s\to y$. Let $s=f(t)$. Using first $(3)$ and $(2)$, we obtain
\begin{align}
h(t)-h(x) &= g(f(t))-g(f(x))\\
          &= [f(t)-f(x)]\cdot[g'(y)+v(s)]\\
          &= (t-x)\cdot[f'(x)+u(t)]\cdot[g'(y)+v(s)],
\end{align}
or, if $t\neq x$,
$$
\frac{h(t)-h(x)}{t-x}=[g'(y)+v(s)]\cdot[f'(x)+u(t)].\tag{4}
$$
Letting $t\to x$, we see that $s\to y$, by the continuity of $f$, so that the right side of $(4)$ tend to $g'(y)f'(x)$, which gives $(1)$. $\Box$
