[FR] The Chain Rule

What is the Chain Rule? Without any knowledge of differential calculus, it will be impossible to explain. However, for those who know derivatives and rules of differentiation, we can start this lecture.

One set of the basic differentition rules is that of sine and cosine.

d/dx [sin(x)] = cos(x) and d/dx [cos(x)] = -sin(x)

But what happens when the argument x has a coefficient? Or what if the argument is a sum or difference?

The first instinct would be to simply leave the argument in place and take the derivative:

d/dx [sin(3x)] = cos(3x)

But this is incorrect.

Remember how derivatives could be described as a rate of change? Consider this example:
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You are given a metal bar of consistant density. You are asked to determine the rate at which the bar expands or contracts due to heat. The metal bar is being cooled at a rate of 5 °C per minute and that the bar expands 2 centimeter per °C.

Using simple dimensional analysis, you can conclude that the bar is expanding at -10 cm/°C. (Negative because the bar is being cooled). So, how did you do it? Well you multiplied the two rates! And that is the foundation of the chain rule.
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Now (hopefully) I've motivated you calculus students to understand that given two rates of change for a single function, the total rate of change can be found by simply multiplying the two rates. Let's refer back to the example:

We know that the rate at which sin(x) changes d/dx [sin(x)] = cos(x). And we know the rate at which 3x changes d/dx 3x = 3.

Keeping the argument consistant, d/dx [sin(3x)] = cos(3x) ∙ 3. The derivative (rate of change) of the entire function is the product of its two individual rates of change.

The actual proof of the Chain Rule is long and complex. But the actual theorem says:
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If g is differentiable at x and f is differentiable at g(x), then the composite function F = f ○ g defined by F(x) = f(g(x)) is differentiable at x and F ' is given by the product

F '(x) = f '(g(x)) ∙ g '(x)

In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, then

 dy/dx = dy/du ∙ du/dx

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One more example will be done with an argument of sums.

Find d/dx [cos(13x + x^2)], where x^2 represents the square of x.

Using the Chain Rule definition:
f (x) = cos(x)          and          g(x) = 13x + x^2
f ' (x) = -sin(x)        and         g '(x) = 13 + 2x

F '(x) = -sin(13x + x^2) ∙ (13 + 2x)