Differentiate #e^(ax)# using first principles?

1 Answer
Dec 12, 2016

# f'(x) = ae^(ax) #

Explanation:

The definition of the derivative of #y=f(x)# is

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So Let # f(x) = e^(ax) # then;

# \ \ \ \ \ f(x+h) = e^(a(x+h)) #
# :. f(x+h) = e^(ax+ah) #
# :. f(x+h) = e^(ax)e^(ah) #

And so the derivative of #y=f(x)# is given by:

# \ \ \ \ \ f'(x) = lim_(h rarr 0) ( (e^(ax)e^(ah)) - (e^(ax)) ) / h #
# :. f'(x) = lim_(h rarr 0) ( e^(ax)(e^(ah) - 1 )) / h #
# :. f'(x) = e^(ax)lim_(h rarr 0) ( (e^(ah) - 1 )) / h #
# :. f'(x) = e^(ax)lim_(h rarr 0) ( a(e^(ah) - 1 )) / (ah) #
# :. f'(x) = ae^(ax)lim_(h rarr 0) ( (e^(ah) - 1 )) / (ah) #

And the clever readers who already know the answer can hopefully spot that we are almost there if we can show that

#lim_(h rarr 0) ( (e^(ah) - 1 )) / (ah) = 1#

Now depending upon how you have defined #e# and the level of calculus that you are at, then this limit can be show to be true. So I wont prove the limit, but just accept it. Wikipedia explains some of the definitions for e using the limit definition.

Once the limit has been established, then the result is evident giving:

# f'(x) = ae^(ax) #