How do you differentiate #y=ln(3x)#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Shwetank Mauria Jul 24, 2016 #(dy)/(dx)=1/x# Explanation: Let #g(x)=3x#, hence #y=y(g(x))#, where #f(x)=lnx# Now using chain rule, #(dy)/(dx)=(dy)/(dg)xx(dg)/(dx)# Hence #(dy)/(dx)=1/(3x)xx3=1/x# Altrnatively #y=ln(3x)=ln3+lnx# and as #ln3# is a constant term #(dy)/(dx)=d/(dx)lnx=1/x# Answer link Related questions What is the derivative of #f(x)=ln(g(x))# ? What is the derivative of #f(x)=ln(x^2+x)# ? What is the derivative of #f(x)=ln(e^x+3)# ? What is the derivative of #f(x)=x*ln(x)# ? What is the derivative of #f(x)=e^(4x)*ln(1-x)# ? What is the derivative of #f(x)=ln(x)/x# ? What is the derivative of #f(x)=ln(cos(x))# ? What is the derivative of #f(x)=ln(tan(x))# ? What is the derivative of #f(x)=sqrt(1+ln(x)# ? What is the derivative of #f(x)=(ln(x))^2# ? See all questions in Differentiating Logarithmic Functions with Base e Impact of this question 11528 views around the world You can reuse this answer Creative Commons License