 | 一問目)
y'=1/√(x^2+1)-x(x+1)/(x^2+1)^(3/2)
=(1-x)/(x^2+1)^(3/2)
∴
y"=-1/(x^2+1)^(3/2)-3x(1-x)/(x^2+1)^(5/2)
=(2x^2-3x-1)/(x^2+1)^(5/2)
となります。
二問目)
y'=1/(1-x^2)^(1/3)+(1/3)(2x^2)/(1-x^2)^(4/3)
=(1/3)(3-x^2)/(1-x^2)^(4/3)
∴
y"=(1/3){(-2x)/(1-x^2)^(4/3)+(4/3)2x(3-x^2)/(1-x^2)^(7/3)}
=(1/9){(-6x)(1-x^2)+8x(3-x^2)}/(1-x^2)^(7/3)
=(2x/9){-3(1-x^2)+4(3-x^2)}/(1-x^2)^(7/3)
=(2x/9)(9-x^2)/(1-x^2)^(7/3)
となります。
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