## 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 du/dx, then the composite function f(u) is differentiable, and

 ddx [f(u)] = f'(u) dudx .

Proof From the definition of the derivative,

 u(x+h) - u(x)h u'(x) as h0.

Thus,

 u(x+h) - u(x)h - u'(x) 0 as h0.

If we take v to be the quantity

 v = u(x+h) - u(x)h - u'(x) ,

then v0 as h0. If we solve this equation for u(x+h), we get

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

where v0 as h0

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

w=
 f(y + k) - f(y)k - f'(y)
ifk 0
0ifk = 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 w0 as k0.

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

 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 h0, 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

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

Now let h0. Since both v and w0 (see above for the reason that w0), we obtain

 limh0 f(u(x+h)) - f(u(x))h = f'(u(x)) u'(x),

which is the chain rule.