求下列函數(shù)的導(dǎo)數(shù)
(1) y=(2x²+3)(3x-1)
y=(2x²+3)(3x-1)=6x³-2x²+9x-3,故y′=18x²-4x+9
(2) y=√(x-2)²
y=√(x-2)²=︱x-2︱
當(dāng)x≧2時,y=x-2,此時y′=1；當(dāng)x≦2時,y=-(x-2)=-x+2,此時y′=-1.
(3) y=x-sin(x/2)cos(x/2)
y′=x′-{[sin(x/2)]′cos(x/2)+sin(x/2)[cos(x/2)]′}
=1-{[cos(x/2)](x/2)′cos(x/2)+sin(x/2)[-sin(x/2)](x/2)′}
=1-[(1/2)cos²(x/2)-(1/2)sin²(x/2)]=1-(1/2)[cos²(x/2)-sin²(x/2)]=1-(1/2)cosx
注“為什么(3x)'=3?”是因為(3x)′=3′x+3x′=0×x+3×1=3
常量的導(dǎo)數(shù)=0,即c′=0,3是常量,故3′=0；(xⁿ)′=nxⁿֿ¹.很好!但時間上還可以再快!