How do you differentiate #y = 1 / log_2 x#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions without Base e 1 Answer Shwetank Mauria Mar 31, 2018 #(dy)/(dx)=-ln2/(x(lnx)^2)# Explanation: #log_2x=lnx/ln2# Therefore #y=ln2/lnx=ln2*1/lnx# and #(dy)/(dx)=ln2*(-1/(lnx)^2)*1/x# = #-ln2/(x(lnx)^2)# Answer link Related questions What is the derivative of #f(x)=log_b(g(x))# ? What is the derivative of #f(x)=log(x^2+x)# ? What is the derivative of #f(x)=log_4(e^x+3)# ? What is the derivative of #f(x)=x*log_5(x)# ? What is the derivative of #f(x)=e^(4x)*log(1-x)# ? What is the derivative of #f(x)=log(x)/x# ? What is the derivative of #f(x)=log_2(cos(x))# ? What is the derivative of #f(x)=log_11(tan(x))# ? What is the derivative of #f(x)=sqrt(1+log_3(x)# ? What is the derivative of #f(x)=(log_6(x))^2# ? See all questions in Differentiating Logarithmic Functions without Base e Impact of this question 1490 views around the world You can reuse this answer Creative Commons License