ответ: f '(0) = f '(3π/2) = 0 .
Объяснение:
f(x)=sin²x ; x=0 ; x=3п/2 ;
f '(x) = ( sin²x )' = 2sinx *( sinx )' = 2sinxcosx = sin2x ;
f '(x) = sin2x ; f '(0) = sin( 2*0) = sin0 = 0 ;
f '(3π/2) = sin(2* 3π/2) = sin3π = sinπ = 0 .
ответ: f '(0) = f '(3π/2) = 0 .
Объяснение:
f(x)=sin²x ; x=0 ; x=3п/2 ;
f '(x) = ( sin²x )' = 2sinx *( sinx )' = 2sinxcosx = sin2x ;
f '(x) = sin2x ; f '(0) = sin( 2*0) = sin0 = 0 ;
f '(3π/2) = sin(2* 3π/2) = sin3π = sinπ = 0 .