How do you find the derivative of #y=a^x#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Jim H Apr 15, 2015 For #a > 0#. If you haven't memorized #d/(dx)(a^x) = a^x lna#, then you use #y=a^x = e^(ln(a^x)) = e^(xlna)# and differentiate using the chain rule to get: #y' = e^(xlna) (lna) = a^x lna# 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 32250 views around the world You can reuse this answer Creative Commons License