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

1. clarence

2. FibonacciChick666

ok, first, find derivative

3. FibonacciChick666

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

4. clarence

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

5. anonymous

you could, its just a bit more difficult that way

6. FibonacciChick666

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

7. anonymous

@Clarence let us know if you are still stuck

8. clarence

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

9. FibonacciChick666

no, first break up the integral. then apply FTC

10. zepdrix

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

11. FibonacciChick666

check out the link, it is really clear

12. anonymous

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

13. anonymous

integrand*

14. 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.

15. 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.

16. FibonacciChick666

Nice explanation zep

17. 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,$\large\rm =\color{orangered}{\cos((}-x^2\color{orangered}{)^2)}(-x^2)'-\color{orangered}{\cos((}\cos x\color{orangered}{)^2)}(\cos x)'$

18. zepdrix

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

19. zepdrix

I should go to bed and stop rambling probably

20. zepdrix

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

21. clarence

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

22. zepdrix

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

23. clarence

Oh boy...

24. zepdrix

Boy this problem is a doozy :p

25. clarence

Yeah, send help :p

26. clarence

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

27. zepdrix

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

28. 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!

29. zepdrix

Hmm, your teacher seems mean -_-

30. clarence

So it'll just be the front bit!

31. clarence

He really is :p

32. zepdrix

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

33. clarence

Thanks! :)