Kakšna je vrednost F '(x), če je F (x) = int_0 ^ sinxsqrt (t) dt?

Odgovor:

#:. F '(x) = (sqrtsinx) (cosx).

Pojasnilo:

#F (x) = int_0 ^ sinx sqrttdt #

# ker, intsqrttdt = intt ^ (1/2) dt = t ^ (1/2 + 1) / (1/2 + 1) = 2 / 3t ^ (3/2) + c, #

#:. F (x) = 2 / 3t ^ (3/2) _ 0 ^ sinx #

#:. F (x) = 2 / 3sin ^ (3/2) x #

#:. F '(x) = 2/3 {(sinx)} ^ (3/2)' #

Uporaba pravila verige, #F '(x) = 2/3 3/2 (sinx) ^ (3 / 2-1) d / dx (sinx) #

# = (sinx) ^ (1/2) (cosx) #

#:. F '(x) = (sqrtsinx) (cosx).

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