[MUSIC]. 
We've already seen how the chain rule can 
help us find the anti-derivatives. 
Alright, that message is called u 
substitution. 
I'm going to do more messing around with 
the chain rule and integration. 
In particular, let's see what we can do 
with the chain rule, and the end points 
of integration. 
Well here's an example to play around 
with, if I take the integral from 0 to x, 
of sine of t dt, well that's something. 
But let's think about the derivative of 
this with respect to x. 
And by the fundamental theorem of 
calculus this is sine of x. 
The fundamental theorem of calculus is 
the derivative of the accumulation 
function is the integrand. 
What if that upper endpoint were a 
function of x? 
What I'm asking is what if this endpoint 
weren't x anymore? 
But some function g of x right? 
Then this wouldn't be sine of x anymore, 
be something else. 
So to analyze this let's define a 
function f. 
So that function will be given by this 
rule. 
f of x is the integral from 0 to x of 
sine t dt. 
Now what would it mean if we replace that 
right hand endpoint instead of being x if 
we replace that with g of x? 
That would mean that I want to think 
about f of g of x, right? 
What's f of g of x? 
So, that's the integral from 0 to g of x, 
of sine t dt. 
And specifically, right, what I want to 
know, is what's the derivative with 
respect x of f of g of x. 
That's just the chain-rule. 
So let me just write down the chain rule. 
The derivative of f of g of x is the 
derivative of f at g of x times the 
derivative of g. 
So what does it mean in this specific 
case? 
Remember in this case, f of x is the 
integral from 0 to x of sine t dt. 
So the derivative of f, by the 
fundamental theorem is sin of x and that 
means that the derivative of f of g of x, 
well that's just sin the derivative of f 
at g of x times the derivative of g. 
Let's make it even more concrete. 
Let's say that g is the squaring 
function. 
Let me write that down. 
So if I'm making g the squaring function, 
then the derivative of g is just 2x and 
in that case the derivative of f of g of 
x is sine of g which is x squared times 
the derivative of g which is 2x. 
So let's write down our final answer in 
this specific case. 
The final claim is that the derivative of 
the integral from 0 to g of x which in 
this case is x squared of sine t dt. 
This is the derivative with respect to x 
of f of g of x. 
This is sin of x squared times 2x and 
this statement, we can get by combining 
the chain rule and the fundamental 
theorem of calculus.