一道高数导数题
1个回答
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)}
