What is the derivative of #log(8x-1)#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions without Base e 1 Answer Bill K. Jul 30, 2015 #8/(ln(10)(8x-1)) approx 3.474/(8x-1)# Explanation: Assuming #log(x)=log_{10}(x)#, #d/dx(log(x))=1/(ln(10)x)#, where #ln(10)=log_{e}(10)# (and #e approx 2.71828#). By the Chain Rule, #d/dx(f(g(x)))=f'(g(x))*g'(x)#, we get #d/dx(log(8x-1))=1/(ln(10)(8x-1)) * d/dx(8x-1)# #=8/(ln(10)(8x-1)) approx 3.474/(8x-1)#. 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 1858 views around the world You can reuse this answer Creative Commons License