Question

I don't understand this question. It says to evaluate the derivative at x = pi and gives the following equation:

Answer 1

ok, first, find derivative

Answer 2

or, better yet, read here: http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doib.html
it explains it way better than me

Answer 3

So I should not find the integral in the bracket first?

Answer 4

you could, its just a bit more difficult that way

Answer 5

^what chris said. DOn't bother finding that first. Split it up, make your life easier

Answer 6

@Clarence let us know if you are still stuck

Answer 7

Okay okay oaky, backing up to the beginning. First, I should find the derivative of cos t^2?

Answer 8

no, first break up the integral. then apply FTC

Answer 9

No.
This is an application of the `Fundamental Theorem of Calculus: Part 1`.\[\large\rm \frac{d}{dx}\int\limits_c^x f(t)dt=f(x)\]

Answer 10

check out the link, it is really clear

Answer 11

the derivative of an integral is always its integral. however, with certain bounds, we can apply some neat stuff to apply this theorem

Answer 12

integrand*

Answer 13

\[\large\rm \frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{\cos(t^2)}dt=\frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{f(t)}dt\]Just generalize the thing that's in the integral.
You "integrate" getting some function, let's call it F(t).\[\large\rm =\frac{d}{dx}~F(t)|_{\cos(x)}^{-x^2}\]Evaluate it at the bounds,\[\large\rm =\frac{d}{dx}\left(F(-x^2)-F(\cos x)\right)\]Then as a final step, you take derivative,
this F turns back into f.\[\large\rm =f(-x^2)(-x^2)'-f(\cos x)(\cos x)'\]We get a bunch of chain rule business going on here.

Answer 14

What's weird about this process is that you're never actually looking for F,
you just assume it exists, and that we'll get back to f by the end of the problem.

~ pseudo anti-differentiate
~ plug in bounds
~ differentiate

You're integrating, then undoing the integration with derivative.
All that happens in the middle is that your variable changes from t to x because of the bounds.

Answer 15

Nice explanation zep

Answer 16

\[\large\rm =\color{orangered}{f(}-x^2\color{orangered}{)}(-x^2)'-\color{orangered}{f(}\cos x\color{orangered}{)}(\cos x)'\]Recall what your original f(t) was,\[\large\rm \color{orangered}{f(}t\color{orangered}{)}=\color{orangered}{\cos((}t\color{orangered}{)^2)}\]The square makes it a little tricky :p hmm...
Place f back where it should be,\[\large\rm =\color{orangered}{\cos((}-x^2\color{orangered}{)^2)}(-x^2)'-\color{orangered}{\cos((}\cos x\color{orangered}{)^2)}(\cos x)'\]

Answer 17

And then they want you to evaluate this at pi? 0_o
hmm

Answer 18

I should go to bed and stop rambling probably

Answer 19

Oh I didn't plug the lower bound in correctly btw, woops\[\large\rm =\color{orangered}{\cos((}-x^2\color{orangered}{)^2)}(-x^2)'-\color{orangered}{\cos((}3\cos x\color{orangered}{)^2)}(3\cos x)'\]

Answer 20

So we just replace the x's with pi's and thus get 3 cos(9) - pi^2 cos(pi^4) right?

Answer 21

Woops, still have chain rule to deal with.
See the primes?
Need to take those derivatives.

Answer 22

Oh boy...

Answer 23

Boy this problem is a doozy :p

Answer 24

Yeah, send help :p

Answer 25

Wait, so what do I have to get at the end of it then?

Answer 26

Sec, Imma back up a sec, make sure I didn't make any boo boos before that

Answer 27

By Fundamental Theorem we get,\[\large\rm =\cos((-x^2)^2)(-x^2)'-\cos((3\cos x)^2)(3\cos x)'\]\[\large\rm =\cos((-x^2)^2)(-2x)-\cos((3\cos x)^2)(-3\sin x)\]Simplifying,\[\large\rm =-2x \cos(x^4)+3\sin x \cos(9\cos^2(x))\]And then ya, plugging in pi,\[\large\rm =-2\pi \cos(\pi^4)+3\sin \pi \cos(9\cos^2(\pi))\]Ooo and that should simplify down further which is nice!

Answer 28

Hmm, your teacher seems mean -_-

Answer 29

So it'll just be the front bit!

Answer 30

He really is :p

Answer 31

ya, that's what it's looking like! :D

Answer 32

Thanks! :)