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

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Math2400
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3what is the fundamental thrm of calculus for integration?

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0\[ \int\limits_a^b f(x)dx=F(b)F(a)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3good, 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?

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0okay got that, and um you get the derivative..? lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3whats it look like tho, we are working a generalization

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0don't u have to undo the antiderivative first on the Right side?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3we already have, its the right side

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3the right side is the 'undoing' of the derivative of the left side. f' comes from f

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0oh okay so if i take the derivative of f(b)f(a)?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3almost, a and b need not be constants, they can be functions, so we apply the chain rule \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=\frac d{dx}f(b)\frac d{dx}f(a)\] \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=b'~f'(b)a'~f'(a)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3notice here, that we already know what f' is ... we know a and b, therefore we know a' and b'

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0ok so u u just need to get the derivatives of a and b and multiy by f'

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\] f'(a) = cos(3(0)+1) f'(b) = cos(3x^5+1) a(x) = 0, a' = 0 b(x) = x^5, b' = 5x^4

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3you 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

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0so when i rewrite it into the equ. i use f'(a) and f'(b) instead of putting the original....so there should be no "t" variable

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3correct, the a and b are in the domain of t in this case.
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