Question

一二题的二阶导数等于多少，求大佬解答？

Answer 1

1.y'=2arctan(x/2)*1/[1+(x/2)^2]*1/2
=4arctan(x/2)/(4+x^2),
y''={4/[1+(x/2)^2]*1/2}/(4+x^2)-4arctan(x/2)*2x/(4+x^2)^2
=[8-8xarctan(x/2)]/(4+x^2)^2.
2.y'=e^[-sin^(1/x)]*[-2sin(1/x)cos(1/x)*(-1/x^2)]
=e^[-sin^(1/x)]sin(2/x)/x^2,
y''=e^[-sin^(1/x)]{[sin(2/x)/x^2]^2+[cos(2/x)*(-2/x^2)]/x^2-2sin(2/x)/x^3}
=e^[-sin^(1/x)]*[sin^(2/x)-2xsin(2/x)-2cos(2/x)]/x^4.
注:sin^u=(sinu)^2.

Answer 2

①y'=2arctan½x·(arctan½x)'
=arctan½x/(1+¼x²)
y''=[½(1+¼x²)/(1+¼x²)+arctan½x·½x]/(1+¼x²)²
②y'=e^[-sin²(1/x)]·[-2sin(1/x)·cos(1/x)·-1/x²]
=sin(2/x)·e^[-sin²(1/x)]/x²
y''={-2cos(2/x)e^[-sin²(1/x)]+x²·sin(2/x)·y'-2xy'}/x⁴