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

- Clarence

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- Clarence

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- FibonacciChick666

ok, first, find derivative

- FibonacciChick666

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

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## More answers

- Clarence

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

- anonymous

you could, its just a bit more difficult that way

- FibonacciChick666

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

- anonymous

@Clarence let us know if you are still stuck

- Clarence

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

- FibonacciChick666

no, first break up the integral. then apply FTC

- zepdrix

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

- FibonacciChick666

check out the link, it is really clear

- anonymous

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

- anonymous

integrand*

- zepdrix

\[\large\rm \frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{\cos(t^2)}dt=\frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{f(t)}dt\]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.

- zepdrix

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.

- FibonacciChick666

Nice explanation zep

- zepdrix

\[\large\rm =\color{orangered}{f(}-x^2\color{orangered}{)}(-x^2)'-\color{orangered}{f(}\cos x\color{orangered}{)}(\cos x)'\]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,\[\large\rm =\color{orangered}{\cos((}-x^2\color{orangered}{)^2)}(-x^2)'-\color{orangered}{\cos((}\cos x\color{orangered}{)^2)}(\cos x)'\]

- zepdrix

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

- zepdrix

I should go to bed and stop rambling probably

- zepdrix

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

- Clarence

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

- zepdrix

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

- Clarence

Oh boy...

- zepdrix

Boy this problem is a doozy :p

- Clarence

Yeah, send help :p

- Clarence

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

- zepdrix

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

- zepdrix

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!

- zepdrix

Hmm, your teacher seems mean -_-

- Clarence

So it'll just be the front bit!

- Clarence

He really is :p

- zepdrix

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

- Clarence

Thanks! :)

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