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

1. anonymous

2. 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 ?

3. campbell_st

well isn't $f(x) = \int\limits f'(x)~ dx$

4. ganeshie8

use FTC + Chain rule

5. anonymous

+ chain rule ?

6. 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)?

7. ganeshie8

Let $$G(t) = \int \cos(t^4)\,dt$$ then do we have $$G'(t) = \cos(t^4)$$ ?

8. 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

9. anonymous

wait wait g(1) would equal $\int\limits \cos(1)$ right?

10. ganeshie8

It doesn't matter, it is just a constat it vanishes when u differentiate

11. 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}

12. anonymous

okay i see

13. ganeshie8

do you see the chain rule part ?

14. anonymous

so $-3x^2\cos (x^{12})$ , yes i do

15. ganeshie8

looks good! here is a general formula : $\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g(x) - h(f(x))*f'(x)$

16. anonymous

okay will take note of that thank you ! :D

17. ganeshie8

corrected a small mistake : $\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g{\color{red}{'}}(x) - h(f(x))*f'(x)$

18. 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