设函数y=tan的平方(1+X的平方),求y的微分Dy.

网友回答

dy=2tan(1+X^2)dtan(1+X^2) =2tan(1+X^2)* 1/cos^2(1+X^2)d(1+X^2) =
2tan(1+X^2)* 1/cos^2(1+X^2)*2xdX =4x*sin(1+X^2)/cos^3(1+X^2) *dx
======以下答案可供参考======
供参考答案1:
y=[tan(1+x^2)]^2
则y'=2tan(1+x^2)·{1/[cos(1+x^2)]^2}·2x
=4xtan(1+x^2)/{[cos(1+x^2)]^2}
所以，微分dy=【4xtan(1+x^2)/{[cos(1+x^2)]^2}】dx