y=根号下x^2+1的微分怎么求.

网友回答

∵y=√(x²+1)
∴dy=d(√(x²+1))
=(√(x²+1))'dx
=(1/2)(x²+1)'/[√(x²+1)]dx
=(1/2)(2x)/[√(x²+1)]dx
=xdx/[√(x²+1)].
======以下答案可供参考======
供参考答案1:
x除以根号下的x^2+1~~~