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[MUSIC]. 

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We've already seen how the chain rule can 
help us find the anti-derivatives. 

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Alright, that message is called u 
substitution. 

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I'm going to do more messing around with 
the chain rule and integration. 

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In particular, let's see what we can do 
with the chain rule, and the end points 

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of integration. 
Well here's an example to play around 

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with, if I take the integral from 0 to x, 
of sine of t dt, well that's something. 

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But let's think about the derivative of 
this with respect to x. 

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And by the fundamental theorem of 
calculus this is sine of x. 

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The fundamental theorem of calculus is 
the derivative of the accumulation 

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function is the integrand. 
What if that upper endpoint were a 

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function of x? 
What I'm asking is what if this endpoint 

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weren't x anymore? 
But some function g of x right? 

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Then this wouldn't be sine of x anymore, 
be something else. 

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So to analyze this let's define a 
function f. 

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So that function will be given by this 
rule. 

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f of x is the integral from 0 to x of 
sine t dt. 

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Now what would it mean if we replace that 
right hand endpoint instead of being x if 

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we replace that with g of x? 
That would mean that I want to think 

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about f of g of x, right? 
What's f of g of x? 

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So, that's the integral from 0 to g of x, 
of sine t dt. 

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And specifically, right, what I want to 
know, is what's the derivative with 

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respect x of f of g of x. 
That's just the chain-rule. 

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So let me just write down the chain rule. 
The derivative of f of g of x is the 

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derivative of f at g of x times the 
derivative of g. 

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So what does it mean in this specific 
case? 

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Remember in this case, f of x is the 
integral from 0 to x of sine t dt. 

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So the derivative of f, by the 
fundamental theorem is sin of x and that 

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means that the derivative of f of g of x, 
well that's just sin the derivative of f 

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at g of x times the derivative of g. 
Let's make it even more concrete. 

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Let's say that g is the squaring 
function. 

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Let me write that down. 
So if I'm making g the squaring function, 

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then the derivative of g is just 2x and 
in that case the derivative of f of g of 

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x is sine of g which is x squared times 
the derivative of g which is 2x. 

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So let's write down our final answer in 
this specific case. 

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The final claim is that the derivative of 
the integral from 0 to g of x which in 

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this case is x squared of sine t dt. 
This is the derivative with respect to x 

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of f of g of x. 
This is sin of x squared times 2x and 

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this statement, we can get by combining 
the chain rule and the fundamental 

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theorem of calculus.