## Math2400 one year ago can someone explain this derivative ? i have the answer but the steps are unclear to me

1. Math2400

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

2. amistre64

what is the fundamental thrm of calculus for integration?

3. Math2400

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

4. amistre64

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?

5. Math2400

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

6. amistre64

whats it look like tho, we are working a generalization

7. Math2400

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

8. Math2400

*left

9. amistre64

we already have, its the right side

10. amistre64

the right side is the 'undoing' of the derivative of the left side. f' comes from f

11. Math2400

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

12. amistre64

yep

13. amistre64

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

14. amistre64

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

15. Math2400

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

16. amistre64

$\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]$ f'(a) = cos(3(0)+1) f'(b) = cos(3x^5+1) a(x) = 0, a' = 0 b(x) = x^5, b' = 5x^4

17. amistre64

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

18. Math2400

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

19. amistre64

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

20. Math2400

i got it actually!

21. Math2400

thank u so much :)

22. amistre64

youre welcome