✔ 最佳答案
1.y"=?(1) y=(1+1/x)^3基本微分公式: (u^n)'=n*u^(n-1)*u'(1/u)'=-u'/u^2(1/u^2)'=-2u'/u^3
y'=3(1+1/x)^2*(1+1/x)'=3(1+1/x)^2*(1/x)'=-3(1+1/x)^2*1/x^2=-3[(1+1/x)/x]^2=-3[(1+x)/x^2]^2=-3(1/x + 1/x^2)^2y"=-6(1/x+1/x^2)*(1/x+1/x^2)'=-6(1/x+1/x^2)(-1/x^2-2/x^3)=6(1/x+1/x^2)(1/x^2+2/x^3)=6(1+1/x)(1+2/x)/x^3.....ans
(2) y=(1-√x)^(-1)y'=-(1-√x)^(-2)*(1-√x)'=-(-√x)'/(1-√x)^2=1/[2√x(1-√x)^2]
1/y'=[2√x(1-√x)^2]
-y"/y'^2=[2√x(1-√x)^2]'=2√x*2(1-√x)(1-√x)'+(2/2√x)*(1-√x)^2=2√x(1-√x)(-√x)'+(1-√x)^2/√x=-√x(1-√x)/√x+(1-√x)^2/√x=(1-√x)(1-√x-√x)/√x-y"=y'^2*(1-√x)(1-2√x)/√xy"=-(1-√x)(1-2√x)/{√x[2√x(1-√x)^2]^2}=-(1-√x)(1-2√x)/{4√x*x(1-√x)^4}=-(1-2√x)/{4√x*x(1-√x)^3}.....ans
2.(1) 錯誤.修改如下:基本微分公式: (u*v)'=uv'+vu'2xy+y^2=x+y2xy'+2y+2yy'=1+y'(2x+2y-1)y'=1-2yy'=(1-2y)/(2x+2y-1).....ans