How do you differentiate #ln(1/(x^2+9))#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Shwetank Mauria Jun 14, 2016 #d/dxln(1/(x^2+9))=-(2x)/(x^2+9)# Explanation: As #f(x)=ln(1/(x^2+9))=ln1-ln(x^2+9)=0-ln(x^2+9)=-ln(x^2+9)# Hence #(df)/(dx)=-1/(x^2+9)xx(2x)=-(2x)/(x^2+9)# 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 4860 views around the world You can reuse this answer Creative Commons License