How do you find the first and second derivative of #y=2^-x#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Astralboy Feb 17, 2017 #f'(x)=-2^(-x)ln2# Explanation: The derivative of any function such as #b^x# is #b^xlnb#. Knowing that, we can clearly see b is 2. But since it's a negative x, we must use chain rule and multiply by #-1# #f(x)=2^(-x)# will become #f'(x)=-2^(-x)ln2# 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 1186 views around the world You can reuse this answer Creative Commons License