1. (xⁿ)'=n*xⁿ⁻¹

а) f'(x) = (8x⁵ - 4x³ + x)' =40x⁴-12x²+1

б) (a/x)'=-a/x²  (√x)'=1/(2√x)  (const)'=0

f'(x) =( 6/x - 7√x - 5)'=-6/x²-7/(2√x)=(-6/x²)-3.5/√x

в) (u*v)'=u'v+uv'

f'(x) = ((5x+3)*(8-4x))'=5*(8-4x)+(5x+3)*(-4)=40-20х-20х-12=28-40х