Question

一道高数导数题

Answer 1

因为：

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

所以：

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

则有：

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

所以重点是求出[x+√(1+x^2)]^(1/3)的导数.

y'

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

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

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

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

=(1/3)*(1+x^2)^(-1/2){[x+√(1+x^2)]^(1/3)+[x+√(1+x^2)]^(-1/3)}

y'(2)=(1/3)*(1+4)^(-1/2){(2+√5)^(1/3)+(2+√5)^(-1/3)}