How do you find the derivative of #x^lnx#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Shwetank Mauria Jun 25, 2016 #(dy)/(dx)=2x^((lnx-1))lnx # Explanation: Let #y=x^(lnx)# hence #lny=lnx xx lnx=(lnx)^2# hence #1/y(dy)/(dx)=2lnx xx 1/x# and #(dy)/(dx)=2lnx xx 1/x xx x^(lnx)=2x^((lnx-1))lnx # Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1114 views around the world You can reuse this answer Creative Commons License