用求复合函数导数的法则:
设y=[x+(x+x^1/2)^1/2]^1/2,
则y'=1/{2[x+(x+x^1/2)^1/2]^1/2}*{1+1/[2(x+x^1/2)^1/2]*[1+1/(2x^1/2)]}
=1/{2[x+(x+x^1/2)^1/2]^1/2}*{1+[2x^1/2+1]/[4x^1/2*(x+x^1/2)]}
=[4x^1/2*(x+x^1/2)+2x^1/2+1]/{8x^1/2*(x+x^1/2)*[x+(x+x^1/2)^1/2]^1/2}.
f’(x)=0.5(x+(x+x^0.5)^0.5)^(-0.5)*(1+0.5(x+x^0.5)^(-0.5))*(1+0.5x^(-0.5))
0.5*(x+(x+x^0.5)^0.5)^(-0.5)*(1+(1+0.5x^(-0.5))*(0.5*(x+x^0.5)^(-0.5)))
1啊