Clarence
  • Clarence
I don't understand this question. It says to evaluate the derivative at x = pi and gives the following equation:
Mathematics
schrodinger
  • schrodinger
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Clarence
  • Clarence
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FibonacciChick666
  • FibonacciChick666
ok, first, find derivative
FibonacciChick666
  • 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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Clarence
  • Clarence
So I should not find the integral in the bracket first?
anonymous
  • anonymous
you could, its just a bit more difficult that way
FibonacciChick666
  • FibonacciChick666
^what chris said. DOn't bother finding that first. Split it up, make your life easier
anonymous
  • anonymous
@Clarence let us know if you are still stuck
Clarence
  • Clarence
Okay okay oaky, backing up to the beginning. First, I should find the derivative of cos t^2?
FibonacciChick666
  • FibonacciChick666
no, first break up the integral. then apply FTC
zepdrix
  • 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
  • FibonacciChick666
check out the link, it is really clear
anonymous
  • 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
  • anonymous
integrand*
zepdrix
  • 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
  • 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
  • FibonacciChick666
Nice explanation zep
zepdrix
  • 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
  • zepdrix
And then they want you to evaluate this at pi? 0_o hmm
zepdrix
  • zepdrix
I should go to bed and stop rambling probably
zepdrix
  • 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
  • Clarence
So we just replace the x's with pi's and thus get 3 cos(9) - pi^2 cos(pi^4) right?
zepdrix
  • zepdrix
Woops, still have chain rule to deal with. See the primes? Need to take those derivatives.
Clarence
  • Clarence
Oh boy...
zepdrix
  • zepdrix
Boy this problem is a doozy :p
Clarence
  • Clarence
Yeah, send help :p
Clarence
  • Clarence
Wait, so what do I have to get at the end of it then?
zepdrix
  • zepdrix
Sec, Imma back up a sec, make sure I didn't make any boo boos before that
zepdrix
  • 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
  • zepdrix
Hmm, your teacher seems mean -_-
Clarence
  • Clarence
So it'll just be the front bit!
Clarence
  • Clarence
He really is :p
zepdrix
  • zepdrix
ya, that's what it's looking like! :D
Clarence
  • Clarence
Thanks! :)

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