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,


If we take $v$ to be the quantity

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

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

Now the same is true for $f:$ Take $w$ to be the quantity

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

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

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

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

Note that, as $h \to 0$, so does the quantity 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

Putting (III) and (V) together now gives

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

Now let $h \to 0$. Since both $v$ and $w \to 0$ (see above for the reason that $w \to 0$), we obtain

which is the chain rule.