✔ 最佳答案
f(x)=2x/(x^2-1)微分兩次怎麼算?
Sol
d[1/(x^2-1)]/dx
=d[1/(x^2-1)]/d(x^2-1)*d(x^2-1)/dx
=-2x/(x^2-1)^2
d[1/(x^2-1)^2]/dx
=d[1/(x^2-1)^2]/d(x^2-1)*d(x^2-1)/dx
=-2/(x^2-1)^3*2x
=-4x/(x^2-1)^3
y=2x/(x^2-1)
y’=2x*(-2x)/(x^2-1)^2+2/(x^2-1)
=-4x^2/(x^2-1)^2+2/(x^2-1)
=-4x^2/(x^2-1)^2+(2x^2-2)/(x^2-1)^2
=(-2x^2-2)/(x^2-1)^2
y"=(-2x^2-2)*(-4x)/(x^2-1)^3+(-4x)/(x^2-1)^2
y”(x^2-1)^3=(-2x^2-2)*(-4x)+(-4x)(x^2-1)
=8x^3+8x-4x^3+4x
=4x^3+12x
y”=(4x^3+12x)/(x^2-1)^3