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anonymous
 one year ago
Find F ′(x) for F(x) = int from x cubed to 1 of (cos(t))^4
anonymous
 one year ago
Find F ′(x) for F(x) = int from x cubed to 1 of (cos(t))^4

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right so i know i have to use the fundamental theory of calculus but usually the end interval is an x so what should i do in this case ?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well isn't \[f(x) = \int\limits f'(x)~ dx\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3use FTC + Chain rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sorry i dont follow , what do i apply the chain rule to, the cos(t^4) or the integral of cos(t^4)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Let \(G(t) = \int \cos(t^4)\,dt \) then do we have \(G'(t) = \cos(t^4)\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[F(x) = \int\limits_{x^3}^1 \cos(t^4)\, dt = G(t) \Bigg_{x^3}^1 = G(1)  G(x^3)\] now differentiate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait wait g(1) would equal \[\int\limits \cos(1)\] right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It doesn't matter, it is just a constat it vanishes when u differentiate

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\begin{align}F'(x) &= \dfrac{d}{dx} \left[G(1)  G(x^3)\right]\\~\\ &=0  G'(x^3) * (x^3)'\\~\\ &=\cos((x^3)^4)*3x^2 \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3do you see the chain rule part ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[3x^2\cos (x^{12})\] , yes i do

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks good! here is a general formula : \[\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g(x)  h(f(x))*f'(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay will take note of that thank you ! :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3corrected a small mistake : \[\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g{\color{red}{'}}(x)  h(f(x))*f'(x)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3well, not small.. but you can see that the derivative kills the integral... you need to plugin the bounds and additionally you also need to multiply the derivative of the bounds
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