What is the derivative of #(3^(2t))/t#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Alan N. Sep 24, 2016 #f'(t) = ((2ln3*t-1)*3^(2t))/t^2# Explanation: #f(t)=3^(2t)/t = 3^(2t)*t^-1# #ln f(t) = 2t*ln3-lnt# Using implicit differentiation: #1/f(t)*f'(t) = 2t*0 + 2*ln3 -1/t# (Product rule and standard differential) #1/f(t)*f'(t) = 2ln3-1/t# Since #f(t)=3^(2t)/t# #f'(t) = 3^(2t)/t * (2ln3-1/t)# #= = ((2ln3*t-1)*3^(2t))/t^2# Answer link Related questions How do I find #f'(x)# for #f(x)=5^x# ? How do I find #f'(x)# for #f(x)=3^-x# ? How do I find #f'(x)# for #f(x)=x^2*10^(2x)# ? How do I find #f'(x)# for #f(x)=4^sqrt(x)# ? What is the derivative of #f(x)=b^x# ? What is the derivative of 10^x? How do you find the derivative of #x^(2x)#? How do you find the derivative of #f(x)=pi^cosx#? How do you find the derivative of #y=(sinx)^(x^3)#? How do you find the derivative of #y=ln(1+e^(2x))#? See all questions in Differentiating Exponential Functions with Other Bases Impact of this question 3491 views around the world You can reuse this answer Creative Commons License