Question

Find dy/dx for y=x^x^x^...^x

Answer 1

Answer:

Step-by-step explanation:

$y=x^{x^{x^{x^{....^{x} } } } } \\y=x^{y} \\lny=ylnx\\\frac{d}{dx}(lny)=\frac{d}{dx}(ylnx)\\\frac{1}{y} \frac{dy}{dx} =\frac{y}{x} +lnx\frac{dy}{dx} \\\\\frac{dy}{dx}=\frac{y^2}{x}+ylnx\frac{dy}{dx}\\\frac{dy}{dx}(1-ylnx)= \frac{y^2}{x}\\\frac{dy}{dx}=\frac{\frac{y^2}{x}}{1-ylnx}\\\frac{dy}{dx}=\frac{y^2}{x(1-ylnx)}$