## Proof of the Chain Rule to accompany Applied Calculus (3e)

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The Chain Rule

If the function $f$ has derivative $f'$ and the function $u$ has derivative $\frac{du}{dx}$, then the composite function $f(u)$ is differentiable, and

 $\frac{d}{dx} [f(u)] = f'(u) \frac{du}{dx}.$

Proof From the definition of the derivative,

$\frac{u(x+h) - u(x)}{h}\ \to u' (x)$ as $h \to 0.$

Thus,

$\frac{u(x+h) - u(x)}{h} - u'(x) \to 0$ as $h \to 0.$

If we take $v$ to be the quantity

$v = \frac{u(x+h) - u(x)}{h} - u' (x),$

then $v \to 0$ as $h \to 0.$ If we solve this equation for $u(x+h),$ we get

$u(x+h) = u(x) + (u'(x) + v)h$     ...   (I)

where $v \to 0$ as $h \to 0$

Now the same is true for $f:$ Take $w$ to be the quantity
$w = \matleft{\* \frac{f(y + k) - f(y)}{k} - f'(y), \quad \text{ if } k \neq 0! 0, \quad \text{ if } k = 0}$

(Note that $w$ is a function of $k$ that is defined for $k$ near, or equal to, $0$). Then

$f(y+k) = f(y) + (f'(y) + w)k$     ...   (II)

whether or not $k = 0,$ and where $w \to 0$ as $k \to 0.$

What we are after is the derivative of $f(u(x)).$ Thus we need to calculate the limit of

$\frac{f(u(x+h)) - f(u(x))}{h}$

First look at the numerator: $f(u(x+h)) - f(u(x)).$ If we use formula (I) to substitute for $u(x+h)$ we get

$f(u(x+h)) - f(u(x)) = f [u(x) + (u'(x) + v)h] - f(u(x))$     ...   (III)

Now we use formula (II) to rewrite $f [u(x) + (u' (x) + v)h]:$

$f [u(x) + (u'(x) + v)h] = f(u(x)) + (f'(u(x)) + w)(u'(x) + v)h$     ...   (IV)
(Take (II) and replace y with, $u(x)$ and $k$ with $(u'(x) + v)h )$

Note that, as $h \to 0$, so does the quantity
$k = (u'(x) + v)h$
in equation (IV), and therefore so does the quantity w by (II). (We will use this fact below.)

Subtracting $f(u(x))$ from both sides of (IV) gives

$f [u(x) + (u'(x) + v)h] - f(u(x)) = (f'(u(x)) + w)(u'(x) + v)h$     ...   (V)

Putting (III) and (V) together now gives

$f(u(x+h)) - f(u(x)) = (f'(u(x)) + w)(u'(x) + v)h$.

This is the numerator of the expression we are after. Dividing by $h$ gives

$\frac{f(u(x+h)) - f(u(x))}{h} = \frac{(f'(u(x)) + w)(u'(x) + v)h}{h} = (f'(u(x)) + w)(u'(x) + v)$  .

Now let $h \to 0$. Since both $v$ and $w \to 0$ (see above for the reason that $w \to 0$), we obtain
$\lim_{h \to 0} \frac{f(u(x+h)) - f(u(x))}{h} = f'(u(x)) u'(x)$,

which is the chain rule.