What is the derivative of #2^x#?

1 Answer
Dec 14, 2015

Answer:

#2^x*ln(2)#

Explanation:

Use the chain rule and the identity:

#d/(dt) e^t = e^t#

Start by using properties of exponents:

#2^x = (e^(ln 2))^x = e^(x ln 2)#

So if we put #t = x ln 2#, then:

#(dt)/(dx) = ln 2#

and:

#d/(dx) 2^x = d/(dx) e^(x ln 2) = (dt)/(dx) d/(dt) e^t = e^t * ln(2)#

#=e^(x ln 2)*ln(2)=2^x*ln(2)#