求y=tan[2x/(1+x^2)]的导数/>

求y=tan[2x/(1+x^2)]的导数/>
求y=tan[2x/(1+x^2)]的导数

/>

求y=tan[2x/(1+x^2)]的导数/>
复合函数求导
y'=sec^2[2x/(1+x^2)]*[2x/(1+x^2)]'
=sec^2[2x/(1+x^2)]*[(2x)'(1+x^2)-2x(1+x^2)]/(1+x^2)^2
=sec^2[2x/(1+x^2)]*[2(1+x^2)-4x^2]/(1+x^2)^2
=sec^2[2x/(1+x^2)]*2(1-x^2)/(1+x^2)^2

y=tan[2x/(1+x^2)]
所以，y'=sec^2 [2x/(1+x^2)]*[2x/(1+x^2)]'
=sec^2 [2x/(1+x^2)]*[(2x)'*(1+x^2)-2x*(1+x^2)']/(1+x^2)^2
=sec^2 [2x/(1+x^2)]*[2(1+x^2)-2x*2x]/(1+x^2)^2
=sec^2 [2x/(1+x^2)]*[2(1-x^2)/(1+x^2)^2]