How do you differentiate #y=ln(1/x)#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Jim H Jul 28, 2016 Simplest way is to rewrite it first. Explanation: #y=ln(1/x)=ln(x^-1) = -1 lnx# So #y' = -1(1/x)= (-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 1206 views around the world You can reuse this answer Creative Commons License