求导！！
3个回答
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(1)
y=(√x+x³+sinx)/x=x^(-1/2)+x²+(sinx/x)
则，y'=[(-1/2)*x^(-3/2)]+2x+(sinx/x)'
=[(-1/2)*x^(-3/2)]+2x+[(cosx*x-sinx)/x²]
(2)
y=(x+1)(x+2)(x+3)
则，y'=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)
=x²+5x+6+x²+4x+3+x²+3x+2
=3x²+12x+11
(3)
y=-sin(x/2)*[1-2cos²(x/4)]
=-sin(x/2)*[-cos(x/2)]
=sin(x/2)*cos(x/2)
=(1/2)sinx
所以，y'=(1/2)cosx
(4)
y=[1/(1-√x)]+[1/(1+√x)]
=[(1+√x)+(1-√x)]/[(1-√x)*(1+√x)]
=2/(1-x)
=2*[(1-x)^(-1)]
所以，y'=2*(-1)*[(1-x)^(-2)]*(-1)=2*[(1-x)^(-2)]=2/(1-x)²
y=(√x+x³+sinx)/x=x^(-1/2)+x²+(sinx/x)
则，y'=[(-1/2)*x^(-3/2)]+2x+(sinx/x)'
=[(-1/2)*x^(-3/2)]+2x+[(cosx*x-sinx)/x²]
(2)
y=(x+1)(x+2)(x+3)
则，y'=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)
=x²+5x+6+x²+4x+3+x²+3x+2
=3x²+12x+11
(3)
y=-sin(x/2)*[1-2cos²(x/4)]
=-sin(x/2)*[-cos(x/2)]
=sin(x/2)*cos(x/2)
=(1/2)sinx
所以，y'=(1/2)cosx
(4)
y=[1/(1-√x)]+[1/(1+√x)]
=[(1+√x)+(1-√x)]/[(1-√x)*(1+√x)]
=2/(1-x)
=2*[(1-x)^(-1)]
所以，y'=2*(-1)*[(1-x)^(-2)]*(-1)=2*[(1-x)^(-2)]=2/(1-x)²
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