00:01
What happens if you write a function as a composite in different ways? do you get the same derivative each time? the chain rule says you should.
00:13
Try it with the functions.
00:15
Find dy d .dx if y equals x to the three halves by using the chain rule with y as a composite of a, y equals u cubed and u equals square root of x.
00:31
And b y equals square root of u and u equals x cubed okay so let's start with a y is u cubed and u is the square root of x um and it's asked well let's find d y d x so d y d x well why isn't exactly a function of x right y is a function of u and u is a function of x.
01:04
So to do this, what we really do is we do d -y -d -u -d -x, right? and that should give us d -y -d -x.
01:20
So d -y -d -u, what is that? well, d -y -d -u, d -y -d -u is the derivative with respect to you.
01:28
That's going to be 3 u squared.
01:33
And similarly, d -y -u.
01:36
U .dx.
01:39
Well, that's the derivative of x to the one -half.
01:42
So that's one -half, x to the minus one -half, or in other words, one over two square root of x.
01:54
So, d -y -d -x is then d -y -d -u, which is 3 -u squared, times d -u -d -x.
02:07
D -u -d -x is 1 over 2 square root of x...