Question

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

Answer 1

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 ?

Answer 2

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

Answer 3

use FTC + Chain rule

Answer 4

+ chain rule ?

Answer 5

im sorry i dont follow , what do i apply the chain rule to, the cos(t^4) or the integral of cos(t^4)?

Answer 6

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

Answer 7

\[F(x) = \int\limits_{x^3}^1 \cos(t^4)\, dt = G(t) \Bigg|_{x^3}^1 = G(1) - G(x^3)\]

now differentiate

Answer 8

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

Answer 9

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

Answer 10

\[\begin{align}F'(x) &= \dfrac{d}{dx} \left[G(1) - G(x^3)\right]\\~\\
&=0 - G'(x^3) * (x^3)'\\~\\
&=-\cos((x^3)^4)*3x^2
\end{align}\]

Answer 11

okay i see

Answer 12

do you see the chain rule part ?

Answer 13

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

Answer 14

looks good! here is a general formula :

\[\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g(x) - h(f(x))*f'(x)\]

Answer 15

okay will take note of that thank you ! :D

Answer 16

corrected a small mistake :

\[\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}~h(t)\,dt ~~=~~ h(g(x))*g{\color{red}{'}}(x) - h(f(x))*f'(x)\]

Answer 17

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