How do you differentiate #9^x#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Alan N. Aug 26, 2016 #ln9*9^x# Explanation: #f(x) = 9^x# #lnf(x) = xln9# #1/f(x)*f'(x) = ln9# (Implicit differentiation) #f'(x) = ln9*f(x)# #= ln9*9^x# 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 6288 views around the world You can reuse this answer Creative Commons License