Differentiate y=a^xusing first principle?

1 Answer
Bio
Feb 19, 2018

#{dy}/{dx} = a^x ln(a)#

Explanation:

First, you have to know the limit

#lim_{h->0} (a^h-1)/h = ln(a)#

Now, for the problem.

#y(x) = a^x#

#y'(x) = lim_{h->0} (y(x+h)-y(x))/h#

#= lim_{h->0} (a^(x+h) - a^x)/h#

#= lim_{h->0} (a^x*a^h - a^x)/h#

#= a^x lim_{h->0} (a^h - 1)/h#

#= a^x ln(a)#

Related questions
Impact of this question
1087 views around the world
You can reuse this answer
Creative Commons License