Proof of the Chain Rule
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Techniques of Differentiation Topic Summary
Review Exercises on Techniques Index of Calculus Proofs Student Home Everything for Calculus |
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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
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Proof From the definition of the derivative,
 h |
 u'(x) as h0. |
Thus,
h |
- | u'(x) | 0 as h0. |
If we take v to be the quantity
| v | = | h |
- | u'(x) | , |
then v0 as h0. If we solve this equation for u(x+h), we get
where v0 as h0
Now the same is true for f: Take w to be the quantity
| w | = |
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(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 w0 as k0.
What we are after is the derivative of f(u(x)). Thus we need to calculate the limit of
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
Now we use formula (II) to rewrite f [u(x) + (u'(x) + v)h]:
Note that, as h0, so does the quantity
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
h |
= | h |
= | (f'(u(x)) + w)(u'(x) + v). |
Now let h0. Since both v and w0 (see above for the reason that w0), we obtain
| lim h 0 |
h |
= | f'(u(x)) u'(x), |
which is the chain rule.
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