Respuesta :

caylus
Hello,

[tex]y=x^{ \frac{1}{x}}* \dfrac{1-ln(x)}{x^2} \\\\ y=x^{ \frac{1}{x} -2}* (1-ln(x))\\\\ ln(y)= (\dfrac{1}{x} -2})*ln(x)+ln(1-ln(x))\\\\ \dfrac{d(ln(x))}{dx} = \dfrac{1}{y} * \dfrac{dy}{dx} \\ =- \frac{1}{x^2}*ln(x)+( \frac{1}{x}-2)* \frac{1}{x} + \frac{1}{1-ln(x)} *(- \frac{1}{x} )\\\\ \dfrac{dy}{dx}=(x^{ \frac{1}{x}}* \dfrac{1-ln(x)}{x^2}) *(\frac{-1}{x^2}*ln(x)+( \frac{1}{x}-2)* \frac{1}{x} + \frac{1}{1-ln(x)} *\frac{-1}{x} )\\\\ ... [/tex]

I let you simplify.