How do you find the derivative of #f(t)=(3^(2t))/t#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Timber Lin Dec 18, 2017 #f'(t) = (3^(2t)(2*ln(3)*t-1))/t^2# Explanation: #f'(t) = (t*d/dt(3^(2t))-3^(2t)*d/dt(t))/t^2# quotient rule #f'(t) = (t*3^(2t)*ln(3)*d/dt(2t)-3^(2t)*1)/t^2# exponent differentiation, chain rule #f'(t) = (t*3^(2t)*ln(3)*2-3^(2t))/t^2# #f'(t) = (3^(2t)(2*ln(3)*t-1))/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 2727 views around the world You can reuse this answer Creative Commons License