 Welcome back to lecture 46 everyone and we're talking about the funnel theorem calculus part one and I wanted to look at two more examples of using the funnel theorem of calculus part one to help us calculate derivatives of integrals In particular We've seen before that the fundamental theorem calculus part one tells us that we take the derivative with respect to x of the integral function a to x f of t dt That this is simply just equal to f of x right here now in order to use this result We do have to have there's a constant a at the bottom of the integral there has to be some variable x That's at the top of the integral right there, and so that's almost the setting we're in right now We do have this constant one that's on the bottom But we have this function x to the fourth that sits on top right there And how do we compensate for this function x the fourth as opposed to a variable and The answer that question is we are going to use the chain rule if we think of the funnel theorem calculus part one as a Derivative rule we can combine it with other derivative rules like the chain rule and so we're going to decompose this function in the following way So we need to take the derivative of two functions Well two functions that are composed with each other What are those two functions going to be well? We have an outer function which we're going to integrate from one to you this secant T dt and so this will represent our outer function and Then we're going to put inside of it an inner function that inner function is the function x to the fourth Is our inner function right here, and you'll notice that if you take x to the fourth and you plug it in for you This will recapture the original function. We have right here And so we want to take the derivative using this chain rule well because of the decomposition here We're going to take the derivative with respect to x to the fourth this outer function one X to the fourth secant T dt But then by the chain rule we have to multiply by the inner derivative as well We have to take the derivative with respect to x this function x to the fourth And so that right there is a fairly painless compensation to do for these For these type of FTC type problems So we're going to have when we take the derivative So so starting this thing over again when we take the derivative of the integral from 1 to x to the fourth secant T dt By the chain rule we're going to take the derivative of the function of the outer function with respect to x to the fourth This will give us secant of x to the fourth Right, so this is our outer derivative. You basically just get the inside function secant But you have to make sure you insert inside of it the upper limit of the integral Of the integral right and then you times it by the inner derivative x to the fourth prime This is your inner derivative right here In which case then your final answer looks like 4x cube that's the derivative x to the fourth times secant of x to the fourth and So if your upper limit is not x, it's not a variable. It's a little bit more complicated than that It doesn't change the calculation too much What you'll do is you'll just take the you'll just take the inside function that that that that's function that is Inside the integral the integrand make sure you plug the upper limit Into the function so we're gonna plug secant the fourth inside for t But then you have to make sure you take the derivative of that inner function as well Let's look at another example of this one. That's a little bit more complicated You'll see that in this situation We're taking the derivative of the integral that ranges from cosine of x to sine of x of the natural log of 1 plus 2x here You'll see that the lower limit is the function cosine of x and the upper limit is the function sine of x How does one deal with this this right here kind of represents the worst case scenario you get when you start taking derivatives of integral functions So remember for FTC We need to have a low we need the lower limit to be a constant and the upper limit to be a variable or in this Case of function. So first what I want us to do is break it up into two integrals. So we're gonna take the derivative of Two pieces. We're gonna insert some middle number into this expression So we can do this in the following way We're gonna take the integral of cosine of x to zero pick your favorite number of zeros typically a good choice here Take the take the integral from cosine of x to zero of the natural log of 1 plus 2t Dt and add it to the integral from zero to sine of x The natural log of 1 plus 2t so we broke this up into two integrals and In and you'll notice that we took this intermediate value zero that sits somewhere in there again It doesn't matter what number you choose just to pick a constant right here I usually like to pick zero So that the forthcoming arithmetic is a little bit easier if there is any But don't worry about that too much just break it up into two pieces Because to apply FTC one you need a constant on the bottom and you know, you know function on top That's the setting that the second integrals in right now The first integrals not quite there the zeros in the wrong spot. So flop those things around You'll end up with the derivative Ddx of The integral you're gonna get the integral from zero to sine of x natural log of 1 plus 2t Dt and then you get minus the integral from zero to cosine of x Natural log of 1 plus 2t so using some properties of integrals We are able to turn this into a problem similar to the previous one. We just saw we can take the two derivatives separately So we need to take So taking our integral zero to sine of x natural log of 1 plus 2t Dt we're gonna take the derivative of that with respect to x and then subtract from that the integral zero to cosine of x natural log of 1 plus 2t Dt don't forget the differential there. We need to take the derivative of that as well So by derivative rules we can take the derivative of these Both parts well like we saw in the previous example when the upper bound is a Function by the chain rule we're gonna plug that function in for t But we have to make sure we also take the derivative of sine So we're gonna end up with the natural log of 1 plus 2 times sine of x And then we have to take the derivative of sine of x And then we subtract from this again by the chain rule with those associated to FTC 1 We plug the upper limit in for the function so we get the natural log of 1 plus 2 cosine of x and then we take the derivative of cosine and So upon doing that the derivative of sine is a cosine So we end up for the first one cosine of x times the natural log of 1 plus 2 sine of x Like so and then the derivative of cosine is a negative sign So we're gonna get a negative negative, which is a positive sine x times the natural log of 1 plus 2 cosine of x And I'm gonna leave the final answer is that I don't think there's really much simplification We're gonna do beyond this point right here, but we can see this kind of represents the worst-case scenario when it comes to taking derivatives of integral functions make sure that if you have a constant on the bottom Whenever you whenever you take an integral derivative integral make sure there's constant on the bottom and Whatever's on the top gets plugged in for x and don't forget your inner derivatives, right? That's all that we need to see and that's it the funnel theorem calculus part 1 is pretty nice It helps us calculate derivatives of these area functions We'll talk about the funnel theorem calculus part 2 in our next lecture 47 stay tuned for that As always if you have any questions feel free to post them in the comments below I will respond to them in a timely manner and love to answer anyone's questions you might have about these videos What have you if you like these videos? Please click like and subscribe if you want to see some more in the future I'll see you next time everyone. Bye