Differentiate e^(ax)eax 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)