How do you differentiate #f(x)= 2^(3x)#? Calculus Basic Differentiation Rules Summary of Differentiation Rules 1 Answer Lovecraft Oct 14, 2015 #f^'(x) = 3ln(2)2^(3x)# Explanation: First we put this in terms of the base #e# exponential #f(x) = e^(3ln(2)x)# We know that #d/dx(e^(kx)) = ke^(kx)#, so for #k = 3ln(2)# we have #f^'(x) = 3ln(2)2^(3x)# Answer link Related questions What is a summary of Differentiation Rules? What are the first three derivatives of #(xcos(x)-sin(x))/(x^2)#? How do you find the derivative of #(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))#? How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#? How do you find the derivative of #y = s/3 + 5s#? What is the second derivative of #(f * g)(x)# if f and g are functions such that #f'(x)=g(x)#... How do you calculate the derivative for #g(t)= 7/sqrtt#? Can you use a calculator to differentiate #f(x) = 3x^2 + 12#? What is the derivative of #ln(x)+ 3 ln(x) + 5/7x +(2/x)#? How do you find the formula for the derivative of #1/x#? See all questions in Summary of Differentiation Rules Impact of this question 1786 views around the world You can reuse this answer Creative Commons License