Question

高数习题如图所示，求详细解答。 用d^2y/dx^2计算

Answer 1

　　解：设y=x/(1+x^2)，则dy/dx=[(1+x^2)-x*2x]/(1+x^2)^2=(1-x^2)/(1+x^2)^2。
　　∴d²y/dx²=[(1-x^2)/(1+x^2)^2]'=(-2x)(3-x^2)/(1+x^2)^3。
　　供参考。

Answer 2

y=x/(x^2+1),那么y'=[(x^2+1)-x*2x]/(x^2+1)^2=(1-x^2)/(x^2+1)^2
y''=dy'/dx=dy'/d(x^2)*d(x^2)/dx
令u=x^2
y''=d[(1-u)/(u+1)^2]/du*du/dx=[((4*u)/(u+ 1)^3 - 3/(u + 1)^2]*2x
=(8*x^3)/(x^2 + 1)^3 - (6*x)/(x^2 + 1)^2

matlab计算结果
>> syms x
>> diff(x/(x^2+1))

ans =

1/(x^2 + 1) - (2*x^2)/(x^2 + 1)^2

>> diff(1/(x^2 + 1) - (2*x^2)/(x^2 + 1)^2)

ans =

(8*x^3)/(x^2 + 1)^3 - (6*x)/(x^2 + 1)^2

>> diff((1-x)/(1+x)^2)

ans =

(2*(x - 1))/(x + 1)^3 - 1/(x + 1)^2

>> diff(1/(x + 1) - (2*x)/(x + 1)^2)

ans =

(4*x)/(x + 1)^3 - 3/(x + 1)^2