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# Worked example: Finding derivative with fundamental theorem of calculus

## Video udskrift

Let's say that we
have capital F of x as being equal to the
definite integral from pi to x of cotangent
squared of t dt. And what we're
curious about finding is the derivative
of this business. We want to figure out what
F prime of x is equal to. Well this is a
direct application of the fundamental
theorem of calculus. This is just going to be
the derivative with respect to x of all of this craziness. So let me just copy
and paste that. So it's just going to be
the derivative with respect to that. And the fundamental
theorem of calculus tells us that this is going to
be equal to just this function, but it's not going to be
a function of t anymore. It's going to be
a function of x. So it's going to be equal
to cotangent squared of x. And you'll often see
problems like this if you go to calculus
competitions or things like that and, or
maybe certain exams. And It's like, oh my god, I
have to take the antiderivative of all of this
business, evaluate it at the different
boundaries, and then I got to take the derivative. No. You just apply the fundamental
theorem of calculus, and it's actually a very
straightforward and a very fast thing to do. Now let's mix it
up a little bit. Let's say that you
had the expression the definite integral from
pi-- instead of from pi to x, let's say it's
from pi to x squared-- and let me write that x
squared in a different color just to make it clear what I'm
doing-- from pi to x squared of cotangent squared of t dt. And you wanted to take the
derivative of this business. So you want to take the
derivative with respect to x of this business. How would you do it? Well the recognition here is
that this is capital-- capital F of x was defined as this. Now instead of an x,
you have an x squared. So this is the exact same
thing as taking the derivative with respect to x of
capital F of-- not x. That would be this. Instead, we have
capital F of x squared. Where we had an x here before,
we now have an x squared. You can verify. If you had capital
F of x squared, every place where you saw the
x here would be an x squared. So it would look
exactly like this. And so we just have to
apply the chain rule. So this is going to be
equal to the derivative of F with respect to x squared. And this is just straight
out of the chain rule. It's going to be F
prime of x squared times the derivative with
respect to x of x squared. This is the derivative of
F with respect to x squared and then times the derivative
of x squared with respect to x. This is just the chain
rule right over here. So what's F prime of x
squared-- the derivative of F with respect to x squared? Well, if you evaluate F prime
at x squared, instead of just being cotangent
squared of x, it's going to be cotangent
squared of x squared. So this part right over here,
this business right over here, is going to be cotangent
squared of x squared. So this is the derivative of all
of this business with respect to x squared, and then you
have to multiply that times the derivative of x squared with
respect to x, which is just 2x. So there you have it. The derivative of this,
all this craziness, is equal to 2x cotangent
squared of x squared.