Question

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

Answer 1

`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)`