(1+x)^1/x的导数
回答
爱扬教育
2022-06-07
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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
解二：链式法则
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)]
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)]