How do you differentiate #log 4^(x^2)#? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions without Base e 1 Answer Shwetank Mauria Oct 18, 2016 #d/(dx)log4^(x^2)=1.2042x# Explanation: #log4^(x^2)=x^2log4# Hence #d/(dx)log4^(x^2)=d/(dx)(x^2log4)# = #log4*d/(dx)x^2# = #2xlog4# = #2x xx0.6021# = #1.2042x# 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 1234 views around the world You can reuse this answer Creative Commons License