How do you find the derivative of #ln sqrt (x^2-4)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer trosk Mar 8, 2016 #d/dx (ln sqrt (x^2-4)) = x/(x^2-4)# Explanation: #ln sqrt (x^2-4) = 1/2 ln (x^2-4)# Hence: #[ln sqrt (x^2-4)]^' = [1/2 ln (x^2-4)]^'# #=1/2 * (x^2-4)^'/(x^2-4) =1/2 * (2x)/(x^2-4) = x/(x^2-4)# 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 9098 views around the world You can reuse this answer Creative Commons License