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Vrz + $2, r =) +rcOs t, s =x + y sin t; du du when x = 1 J = 2,t = 0 dx dy...

Question

Vrz + $2, r =) +rcOs t, s =x + y sin t; du du when x = 1 J = 2,t = 0 dx dy

Vrz + $2, r =) +rcOs t, s =x + y sin t; du du when x = 1 J = 2,t = 0 dx dy



Answers

Find $ dy/dx $,

$ x = te^1 $, $ \quad y = t + \sin t$

Okay, so we're gonna be using part of the definition of the fundamental theorem of calculus, which I've written up in the top right here, um to find the value of this equation down here. And so basically what the fundamental theorem of calculus tells us um in this context is that when we're taking the derivative of an integral from a constant to I, um variable X, which is the same variable that we're taking the derivative with respect to of some function with a different variable. Um and then that that is going to be equal to F. Of X. So really what we need is we need a constant as our A and then an X variable as R. B. As in we're going to integrate from A to B. We need X equally be an eight equals some constant. And then we just need to be taking the derivative of that integral. And we can go ahead and use this equation up here. And so in this case we can and so we're just gonna say this is equal to F of X. If f f T is equal to sine square T, then f of X is going to be equal to sine squared X. And so this is the value of this derivative

One day we're going to start pretty indecorous zero but after the are equal integral zero but X squared, right, Plus ex way did in Duke minus toe science t I plus Tau Costea into the equal toe We get, like, 02 point minus to take the square scientific this bellick flee cause the the here alone are equal toe cause being I plus toe side the that's it Execute toe to cost the by equal to toe saying he integral zero toe fight minus 24 cause quality Thank t plus doing the four costea scientists Costea the Intel zero toe by minus eight cause quality 70 plus eight, of course quality sci fi duty equal to zero. Thank you.

So here we have y equals in a row from zero to sign in verse of T x. I mean co sign t 80. So they asked us to take the derivative derivative cancels out an integral on Lee thing we need to worry about. It's the chain roll. So you have the coastline of the inverse sine of acts times the derivative of the inverse sine, which is one over square root of one minus X squared, and that looks good to me.


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