Question

can someone explain this derivative ? i have the answer but the steps are unclear to me

Answer 1

\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\]

Answer 2

what is the fundamental thrm of calculus for integration?

Answer 3

\[ \int\limits_a^b f(x)dx=F(b)-F(a)\]

Answer 4

good, but lets use different notation, f= f', and F = f

this helps to see the relationship between F and f in a more normal fashion.

\[\int_{a}^{b}f'(x)~dx=f(b)-f(a)\]

same setup, just more familiar notations

what happens when we take the derivative of both sides?

Answer 5

okay got that, and um you get the derivative..? lol

Answer 6

whats it look like tho, we are working a generalization

Answer 7

don't u have to undo the antiderivative first on the Right side?

Answer 8

*left

Answer 9

we already have, its the right side

Answer 10

the right side is the 'undoing' of the derivative of the left side.

f' comes from f

Answer 11

oh okay so if i take the derivative of f(b)-f(a)?

Answer 12

yep

Answer 13

almost, a and b need not be constants, they can be functions, so we apply the chain rule
\[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=\frac d{dx}f(b)-\frac d{dx}f(a)\]
\[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=b'~f'(b)-a'~f'(a)\]

Answer 14

notice here, that we already know what f' is ... we know a and b, therefore we know a' and b'

Answer 15

ok so u u just need to get the derivatives of a and b and multiy by f'

Answer 16

\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\]

f'(a) = cos(3(0)+1)
f'(b) = cos(3x^5+1)

a(x) = 0, a' = 0
b(x) = x^5, b' = 5x^4

Answer 17

you have all the parts required to determine the results, without havig to intgrate first, and then take the derivative ... its a shortcut if you will

Answer 18

so when i re-write it into the equ. i use f'(a) and f'(b) instead of putting the original....so there should be no "t" variable

Answer 19

correct, the a and b are in the domain of t in this case.

Answer 20

i got it actually!

Answer 21

thank u so much :)

Answer 22

youre welcome