How do you differentiate: #ln(x^2 +1)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer GiĆ³ Jul 22, 2015 I found: #(2x)/(x^2+1)# Explanation: I would use the Chain Rule to dela with #ln# first and multiply times the derivative of the argument to get: #(dy)/(dx)=1/(x^2+1)*2x=(2x)/(x^2+1)# 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 32717 views around the world You can reuse this answer Creative Commons License