How do you differentiate #2^(2x)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Bdub Apr 17, 2016 #d/dx = 2^(2x) ln 2 *2# #=2^(2x+1) ln 2# Explanation: Use formula #b^u= b^u ln b xx u'# #d/dx = 2^(2x) ln 2 *2# #=2*2^(2x) ln2# #=2^(2x+1) ln 2# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1729 views around the world You can reuse this answer Creative Commons License