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Clarence

  • one year ago

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

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  1. clarence
    • one year ago
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  2. FibonacciChick666
    • one year ago
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    ok, first, find derivative

  3. FibonacciChick666
    • one year ago
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    or, better yet, read here: http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doib.html it explains it way better than me

  4. clarence
    • one year ago
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    So I should not find the integral in the bracket first?

  5. anonymous
    • one year ago
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    you could, its just a bit more difficult that way

  6. FibonacciChick666
    • one year ago
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    ^what chris said. DOn't bother finding that first. Split it up, make your life easier

  7. anonymous
    • one year ago
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    @Clarence let us know if you are still stuck

  8. clarence
    • one year ago
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    Okay okay oaky, backing up to the beginning. First, I should find the derivative of cos t^2?

  9. FibonacciChick666
    • one year ago
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    no, first break up the integral. then apply FTC

  10. zepdrix
    • one year ago
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    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
    • one year ago
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    check out the link, it is really clear

  12. anonymous
    • one year ago
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    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
    • one year ago
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    integrand*

  14. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{\cos(t^2)}dt=\frac{d}{dx}\int\limits_{\cos(x)}^{-x^2}\color{orangered}{f(t)}dt\]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
    • one year ago
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    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
    • one year ago
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    Nice explanation zep

  17. zepdrix
    • one year ago
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    \[\large\rm =\color{orangered}{f(}-x^2\color{orangered}{)}(-x^2)'-\color{orangered}{f(}\cos x\color{orangered}{)}(\cos x)'\]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
    • one year ago
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    And then they want you to evaluate this at pi? 0_o hmm

  19. zepdrix
    • one year ago
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    I should go to bed and stop rambling probably

  20. zepdrix
    • one year ago
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    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
    • one year ago
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    So we just replace the x's with pi's and thus get 3 cos(9) - pi^2 cos(pi^4) right?

  22. zepdrix
    • one year ago
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    Woops, still have chain rule to deal with. See the primes? Need to take those derivatives.

  23. clarence
    • one year ago
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    Oh boy...

  24. zepdrix
    • one year ago
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    Boy this problem is a doozy :p

  25. clarence
    • one year ago
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    Yeah, send help :p

  26. clarence
    • one year ago
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    Wait, so what do I have to get at the end of it then?

  27. zepdrix
    • one year ago
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    Sec, Imma back up a sec, make sure I didn't make any boo boos before that

  28. zepdrix
    • one year ago
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    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
    • one year ago
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    Hmm, your teacher seems mean -_-

  30. clarence
    • one year ago
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    So it'll just be the front bit!

  31. clarence
    • one year ago
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    He really is :p

  32. zepdrix
    • one year ago
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    ya, that's what it's looking like! :D

  33. clarence
    • one year ago
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    Thanks! :)

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