^1/x的导数

解一：对数求导法

y = (1+x)^(1/x)

lny = (1/x)ln(1+x)

y'*1/y = ln(1+x)*(-1/x) + (1/x)*1/(1+x)

= (1/x) * [1/(1+x) - (1/x)ln(1+x)]

y' = (1/x)(1+x)^(1/x) * [1/(1+x) - (1/x)ln(2+x)]

扩展资料

y = (1+x)^(1/x),令a = 1+x,z = 1/x

∴y = a^z

dy/dx = d(a^z)/d(a) * d(a)/d(x) + d(a^z)/d(z) * d(z)/d(x)

= (z)a^(z-1) * (0+1) + (a^z)(lna) * (-1/x)

= (z)(a^z)/(a) - (a^z)(lna)(1/x)

= (a^z) * [z/a - (lna)/x]

= (1+x)^(1/x) * [(1/x)/(1+x) - (1/x)ln(1+x)]

= (1/x)(1+x)^(1/x) * [1/(1+x) - (1/x)ln(2+x)]