# Second fundamental theorem and chain rule | MIT 18.01SC Single Variable Calculus, Fall 2010

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05:04   |   Jan 07, 2011

### Transcription

• PROFESSOR: Welcome back to recitation.
• In this video segment, what I'd like us to do
• is work on this following problem.
• Find d/dx of the integral from 0 to x squared, cosine t dt.
• I'm going to give you a moment to think about it
• and then I'll come back and show you how I do it.
• OK, welcome back.
• Hopefully you were able to make some headway on this.
• Let's look at the problem and see how we would break it down.
• Well we know from the fundamental theorem of calculus
• that you saw in the lecture, that if up here
• then the problem would be easy to solve.
• We'd just use the fundamental theorem of calculus, the answer
• would be cosine x.
• But of course, we don't have an x, we have an x squared.
• That's why I gave you this problem.
• And we need to figure out how to solve this problem when
• there's a different function up here besides just x.
• What we're going to use is we're going
• to combine the fundamental theorem of calculus
• and the chain rule.
• So, let's start off with how we would do this
• if it were the integral from 0 to x, as I mentioned.
• So we'll define capital F of x to be
• equal to the integral from 0 to x cosine t dt.
• And then we know that f prime of x is equal to cosine x.
• Now the problem is, we don't just have this, as I mentioned.
• What we actually have-- let me write this down--
• we have F of x squared.
• Right?
• That's what this-- sorry let me highlight
• what I mean-- this boxed thing is F of x squared.
• So we took F of x and we evaluated it at x squared.
• That's what we get in the box.
• And so if we want to find d/dx of F of x squared,
• it really is just the chain rule.
• We really want to think of this as a composition of functions.
• The first, the outside function is capital F
• and the inside function is x squared.
• So just in general, how do we think about the chain rule?
• Well remember what we do-- let me
• come back here for a second-- remember what
• we do is we take the derivative of the outside function,
• we evaluate it at the inside function
• and then we take the derivative of the inside function
• and we multiply those two together.
• So all I have to do is figure out,
• what is the following thing?
• We know d/dx the quantity F of x squared
• should be equal to F prime evaluated
• at x squared times 2x.
• That's just what we said earlier, right?
• It's the derivative of F evaluated
• at x squared times the derivative of x squared.
• So now I just have to figure out what this is.
• Well let's go back to the other side
• and see what we wrote that F prime was.
• If we come over here, we see F prime at x is just cosine x.
• So F prime at x squared is going to be this function evaluated
• at x squared.
• That's just cosine of x squared.
• So we see over here, we just get cosine x squared times 2x.
• And because I'm a mathematician, I
• want to write the 2x in front before I finish.
• Because otherwise I get confused.
• So the answer here is just 2x times cosine x squared.
• Now I want to point out really what we did here.
• This is the answer to this particular problem,
• but we can now solve problems in general,
• when I put any function up here, any function of x up here.
• Right?
• Ultimately, all I did was I used the fundamental theorem
• of calculus and the chain rule.
• So any function I put up here, I can
• do exactly the same process.
• I would define F of x to be this type of thing,
• the way we would define it for the fundamental theorem
• of calculus.
• I would know what F prime of x was.
• And then I would have to evaluate F
• at a, at this function up here, whatever I put up there.
• So in this case it was x squared.
• I could have made it natural log x.
• I could've made it some big polynomial or something more
• complicated.
• Right?
• And once I do that, I just follow this same process.
• Now, instead of the x squared here
• I would have that other function.
• So I'd evaluate capital F at whatever function
• that is times the derivative of that function.
• It's exactly the same process.
• So I want to point out that this is a bigger situation than I
• had before, or a bigger situation than just
• this little problem.
• So, just so you understand that.
• OK?
• So again, I just want to say one more time.
• Now you know how to solve problems
• where you have any other function of x up here
• and you want to take the derivative
• of this kind of expression of an integral
• with another function of x up there.
• All right, I think I'll stop there.

### Description

Second fundamental theorem and chain rule
Instructor: Christine Breiner
View the complete course: http://ocw.mit.edu/18-01SCF10

More courses at http://ocw.mit.edu

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