(You never thought that we would be talking about chains in this class!! J )

The whole purpose of the chain rule is to be able to find the derivative of a complicated function without to much stress. Consider the following composition of functions: p (x) = f (g (k (h (x)))). The derivative p' (x) = f ' (g (k (h (x))))* g' (k (h (x)))* k' (h (x))* h' (x). I know that this is the extreme, but basically this is what you are doing. It is like peeling an orange from the inside out. Take the derivative of the inside, times the derivative of the next layer. Keep doing this until you get to the final layer.

EXAMPLE 1:

Find f ' for f (x) = (x ⁴ + 2x ² - 6) ¹⁰⁰.

SOLUTION:

Remember the rule for f (x) = x ⁿ. f ' (x) = n x ^{n - 1}. n can be any real number, even a fraction.

Notice, that the final answer is the derivative of the inside times the derivative of the outside. Using the chain rule is like peeling an orange inside out. Start with the innermost function and take its derivative, then the next innermost function, and so on.

EXAMPLE 2:

SOLUTION:

Use the product rule to find u'. (f (x) = h (x) * g (x). f ' (x) = h' (x)* g (x) + h (x)* g' (x).)

EXAMPLE 3:

SOLUTION:

EXAMPLE 4:

Find f ' for f (x) = sin (cos (x ² - 5x + 6)).

Try to figure out the steps that I skipped. Remember f ' (p) = f ' (p (u)).

EXAMPLE 5:

You might want to simplify the above result, but it is not necessary.

EXAMPLE 6:

Find f ' for f (x) = (1 + cos 2x) ^{- 4}.

EXAMPLE 7:

Find f ' for f (x) = (csc x + cot x) ^{- 1}.

EXAMPLE 8:

Find f ' for f (x) = sin (cos (2x - 5)).