一道高数导数题
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一道高数导数题
![](http://img.wesiedu.com/upload/6/e2/6e22f3ee3ff779c9ff3b83698d0173be.jpg)
![](http://img.wesiedu.com/upload/6/e2/6e22f3ee3ff779c9ff3b83698d0173be.jpg)
![一道高数导数题](/uploads/image/z/7417742-14-2.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E5%AF%BC%E6%95%B0%E9%A2%98%26nbsp%3B)
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)}
[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)}