Найти f'(x):
F(x)= cos (3-4x);
F(x)= tg (2x-7)
F(x)= sinx * cos (2x-3)

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

F'(x) =(cos(3-4x)) ' =  (-sin(3-4x))*(3-4x)' = (-sin(3-4x))*(0-4) = 4sin(3-4x)   * * * -4sin(4x-3)* * *--------F'(x) =(cos(3-4x)) '  =(cos(4x-3)) ' =  -sin(4x-3)*(4x-3)' = -sin(4x-3)*(4-0)' = -4sin(4x-3).***********************F'(x) =(tq(2x-7))' =(1/cos²(2x-7)) *(2x-7)' = (1/cos²(2x-7)) *(2-0) = 2/cos²(2x-7).***********************F'(x) =(sinx *cos(2x-3) )'  =(sinx)' *cos(2x-3)+sinx* (cos(2x-3) )' =cosx*cos(2x-3) +sinx*(-(sin(2x-3))*(2x-3)' =cosx*cos(2x-3) - sinx*sin(2x-3)*(2-0) =cosx*cos(2x-3) -2 sinx*sin(2x-3).