Hello class, we
are going to learn about the chain rule in today’s class. How exciting, this is
one of my favorite topics!
The chain rule
is one of the most important concepts in calculus, specifically
differentiation. This rule explains how you take the derivative of a
"composition of functions", if both are defined in terms of a certain
variable.
Suppose you have two functions, f(x) and g(x). The
composition of these functions could be:
f(g(x)) or g(f(x))
So what the chain rule says is that the derivative of
f(g(x)) = f ' (g(x)) * g'(x), where f'(x) is the derivative of f with respect
to x, and g'(x) is the derivative of g with respect to x.
So let's demonstrate this idea in several examples:
Example 1) ln (x^2). This is a composition of two functions, ln x and
x^2.
So the outer function in this case, f(x) is equal to ln x. Thus f'(x) = 1/x.
The inner function in this case, g(x) is x^2, and g'(x) = 2x.
So the outer function in this case, f(x) is equal to ln x. Thus f'(x) = 1/x.
The inner function in this case, g(x) is x^2, and g'(x) = 2x.
So f(g(x)) has derivative f'(g(x)) * g'(x).
f'(g(x)) = f'(x^2) = 1/x^2
g'(x) = 2x
g'(x) = 2x
So the product of f'(g(x)) * g'(x) = 1/x^2 * 2x = 2/x
Example 2) Derivative of (sin x)^2
So here the outer function, f(x) is x^2, which has
derivative 2x
and the inner function, g(x), is sin x, which has derivative cos x
and the inner function, g(x), is sin x, which has derivative cos x
so f(g(x)) has derivative f'(g(x)) * g'(x) = f'(sin x) *cos x
= 2sin x cos x
Example 3) Derivative of tan(ln x^2))
This is an example of a composition of more than 2
functions. The three functions in this case are:
f(x) = tan x, so f'(x) = (sec x)^2
g(x) = ln x, g'x) = 1/x
h(x) = x^2, h'(x) = 2x
g(x) = ln x, g'x) = 1/x
h(x) = x^2, h'(x) = 2x
Although there are three functions in this case you still
apply the same rules: the derivative of the outer function applied to the inner
function, times the derivative of the inner function.
So f(g(h(x))) has derivative f'(g(h(x)) * g'(h(x)) * h'(x)
=(sec(g(h(x)))^2 * 1 / h(x) * 2x =
(sec(ln x^2))^2 * 1/x^2 * 2x = (sec(ln x^2))^2 * 2/x
These are just the basics about using the chain rule. You
just need to find which functions form the composition of the function, find
the outer function's derivative, apply it to the inner function, then the inner
function's derivative etc etc etc.
Okay class, I hope you enjoyed this lecture as much as I
did. Its almost final time, keep up the hard work, you guys will do great!!
Awesome work! Very clear, and the diversity in examples is appreciated (how often do you see trig?). I'm glad that you didn't choose to do run-off the mill examples.
ReplyDeletei am glad she chose some meaty examples, also!
DeleteI like how you used several examples to demonstrate the chain rule. I also like how each example was supplemented with worded explanations. Good work!
ReplyDeleteGood job! I like your examples. It was very clear and I understood it.
ReplyDeletelindsey,
ReplyDeletenice explanation of the chain rule and i agree with everyone else that it was nice to see some examples that were a little more hearty!
professor little