What is the derivative of #2^x#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Shwetank Mauria Mar 17, 2018 Derivative of #2^x# is #2^xln2# Explanation: Let #y=2^x#. Now to find #(dy)/(dx)#, let us take logarithm on both sides, which gives #lny=xln2# - now taking implicit differential on both sides #1/y(dy)/(dx)=ln2# and hence#(dy)/(dx)=yln2=2^xln2# 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 1484 views around the world You can reuse this answer Creative Commons License