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)}

Answer 2

令 t=x+√(1+x²)，则 y=t^(¹/³)+[1/t^(¹/³)]，t'=1+[x/√(1+x²)]；
t'(2)=1+[2/√(1+2²)]=1+(2√5/5)，t(2)=2+√5，1/t^(²/³)=1/(2+√5)^(²/³)=(√5-2)^(²/³)；
y'=[(1/3)*t^(-²/³)-(1/3)*t^(-4/3)]*t'={1/[3t^(²/³)] -1/[3t^(4/3)]}*t'；
y'(2)={(1/3)(√5-2)^(²/³)-(1/3)(9-4√5)^(²/³)}*[1+(2√5/5)]；