Question

求帮忙解答一道数学题

已知图1，化简图2，求求了

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Answer 1

y=1/√(x^2+1)
y'=-x/(x^2+1)^(3/2)
y''
=[-(x^2+1)^(3/2) + x. (3/2)(x^2+1)^(1/2). (2x) ]/(x^2+1)^3

=[-(x^2+1)^(3/2) + 3x^2.(x^2+1)^(1/2) ]/(x^2+1)^3
=[-(x^2+1) + 3x^2 ]/(x^2+1)^(5/2)
=(2x^2-1)/(x^2+1)^(5/2)
y''+y'+xy^3

=(2x^2-1)/(x^2+1)^(5/2) -x/(x^2+1)^(3/2) +x[1/√(x^2+1)]^3
=(2x^2-1)/(x^2+1)^(5/2) -x/(x^2+1)^(3/2) +x/(x^2+1)^(3/2)
=(2x^2-1)/(x^2+1)^(5/2)

Answer 2

y=1/√(x^2+1)=(x^2+1)^(-1/2),
y'=(-1/2)(x^2+1)^(-3/2)*2x
=-x(x^2+1)^(-3/2),
y''=-(x^2+1)^(-3/2)+(3x/2)(x^2+1)^(-5/2)*2x
=(2x^2-1)(x^2+1)^(-5/2).
所以y''+y'+xy^3
=(2x^2-1)(x^2+1)^(-5/2)-x(x^2+1)^(-3/2)+x(x^2+1)^(-3/2)
=(2x^2-1)(x^2+1)^(-5/2)
=(2x^2-1)/√(x^2+1)^5.