How do you find the first and second derivative of #lnx^2 #? Calculus Graphing with the Second Derivative Notation for the Second Derivative 1 Answer The_tutor Oct 10, 2016 #y'=2x/x^2# #y''=2/x^2# Explanation: Using the chain rule we get: #y'=(1/x^2)*2x#=#2x/x^2# #y''=((x^2*2)-4x^2)/x^4#=#(2x^2-4x^2)/x^4#=#-2x^2/x^4#=#2/x^2# Answer link Related questions What is notation for the Second Derivative? What is Leibniz notation for the second derivative? What is the second derivative of #e^(2x)#? How do you find the first, second derivative for #3x^(2/3)-x^2#? What is the second derivative of #y=x*sqrt(16-x^2)#? How do you find the first and second derivative of #(lnx)/x^2#? How do you find the first and second derivative of #lnx^(1/2)#? How do you find the first and second derivative of #x(lnx)^2#? How do you find the first and second derivative of #ln(x^2-4)#? How do you find the first and second derivative of #ln(lnx^2)#? See all questions in Notation for the Second Derivative Impact of this question 1771 views around the world You can reuse this answer Creative Commons License