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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:
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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FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2ok, first, find derivative

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2or, better yet, read here: http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doib.html it explains it way better than me

clarence
 one year ago
Best ResponseYou've already chosen the best response.0So I should not find the integral in the bracket first?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you could, its just a bit more difficult that way

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2^what chris said. DOn't bother finding that first. Split it up, make your life easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Clarence let us know if you are still stuck

clarence
 one year ago
Best ResponseYou've already chosen the best response.0Okay okay oaky, backing up to the beginning. First, I should find the derivative of cos t^2?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2no, first break up the integral. then apply FTC

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5No. 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
 one year ago
Best ResponseYou've already chosen the best response.2check out the link, it is really clear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the derivative of an integral is always its integral. however, with certain bounds, we can apply some neat stuff to apply this theorem

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm \frac{d}{dx}\int\limits_{\cos(x)}^{x^2}\color{orangered}{\cos(t^2)}dt=\frac{d}{dx}\int\limits_{\cos(x)}^{x^2}\color{orangered}{f(t)}dt\]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
 one year ago
Best ResponseYou've already chosen the best response.5What'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 antidifferentiate ~ 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
 one year ago
Best ResponseYou've already chosen the best response.2Nice explanation zep

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm =\color{orangered}{f(}x^2\color{orangered}{)}(x^2)'\color{orangered}{f(}\cos x\color{orangered}{)}(\cos x)'\]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
 one year ago
Best ResponseYou've already chosen the best response.5And then they want you to evaluate this at pi? 0_o hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I should go to bed and stop rambling probably

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0So we just replace the x's with pi's and thus get 3 cos(9)  pi^2 cos(pi^4) right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Woops, still have chain rule to deal with. See the primes? Need to take those derivatives.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Boy this problem is a doozy :p

clarence
 one year ago
Best ResponseYou've already chosen the best response.0Wait, so what do I have to get at the end of it then?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Sec, Imma back up a sec, make sure I didn't make any boo boos before that

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5By 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
 one year ago
Best ResponseYou've already chosen the best response.5Hmm, your teacher seems mean _

clarence
 one year ago
Best ResponseYou've already chosen the best response.0So it'll just be the front bit!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5ya, that's what it's looking like! :D
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