How do you find the second derivative of #f(x)=lnx+sqrt(162)#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Massimiliano Mar 1, 2015 The answer is: #y''=-1/x^2#. Since #sqrt162# is a number, so: #y'=1/x=x^-1# and #y''=-1x^(-1-1)=-1/x^2#. 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 1735 views around the world You can reuse this answer Creative Commons License