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

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

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
  • instead of x squared we had an x,
  • 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.

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Description

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

License: Creative Commons BY-NC-SA
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