anonymous
  • anonymous
Find F ′(x) for F(x) = int from x cubed to 1 of (cos(t))^4
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
right 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
  • campbell_st
well isn't \[f(x) = \int\limits f'(x)~ dx\]

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ganeshie8
  • ganeshie8
use FTC + Chain rule
anonymous
  • anonymous
+ chain rule ?
anonymous
  • anonymous
im sorry i dont follow , what do i apply the chain rule to, the cos(t^4) or the integral of cos(t^4)?
ganeshie8
  • ganeshie8
Let \(G(t) = \int \cos(t^4)\,dt \) then do we have \(G'(t) = \cos(t^4)\) ?
ganeshie8
  • ganeshie8
\[F(x) = \int\limits_{x^3}^1 \cos(t^4)\, dt = G(t) \Bigg|_{x^3}^1 = G(1) - G(x^3)\] now differentiate
anonymous
  • anonymous
wait wait g(1) would equal \[\int\limits \cos(1)\] right?
ganeshie8
  • ganeshie8
It doesn't matter, it is just a constat it vanishes when u differentiate
ganeshie8
  • ganeshie8
\[\begin{align}F'(x) &= \dfrac{d}{dx} \left[G(1) - G(x^3)\right]\\~\\ &=0 - G'(x^3) * (x^3)'\\~\\ &=-\cos((x^3)^4)*3x^2 \end{align}\]
anonymous
  • anonymous
okay i see
ganeshie8
  • ganeshie8
do you see the chain rule part ?
anonymous
  • anonymous
so \[-3x^2\cos (x^{12})\] , yes i do
ganeshie8
  • ganeshie8
looks 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
  • anonymous
okay will take note of that thank you ! :D
ganeshie8
  • ganeshie8
corrected 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
  • ganeshie8
well, 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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