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

##### 1 Answer

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

#### Explanation:

The definition of the derivative of

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

So Let

# \ \ \ \ \ 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

# \ \ \ \ \ 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

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

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