If y = x x and x > 0 then ln y = ln (x x) Use properties of logarithmic functions to expand the right side of the above equation as follows. ln y = x ln x We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right. y '(1 / y) = ln x + x(1 / x) = ln x + 1 , where y ' = dy/dx Multiply both sides by y y ' = (ln x + 1)y Substitute y by x x to obtain y ' = (ln x + 1)x x