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

Question

Emma Aldridge · Accepted Answer

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

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) #

Samantha Adkison · Answer

undefined To differentiate $e^{ax}$ using first principles, we use the definition of the derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

For $f(x) = e^{ax}$, the derivative $f'(x)$ is:

$$f'(x) = \lim_{h \to 0} \frac{e^{a(x+h)} - e^{ax}}{h}$$

$$= \lim_{h \to 0} \frac{e^{ax}e^{ah} - e^{ax}}{h}$$

$$= \lim_{h \to 0} \frac{e^{ax}(e^{ah} - 1)}{h}$$

Using the limit definition of the exponential function $e^u$:

$$= \lim_{h \to 0} \frac{e^{ax}(1 + ah + \frac{(ah)^2}{2!} + ...) - e^{ax}}{h}$$

$$= \lim_{h \to 0} \frac{e^{ax} + ae^{ax}h + \frac{(ae^{ax}h)^2}{2!} + ... - e^{ax}}{h}$$

$$= \lim_{h \to 0} \frac{ae^{ax}h + \frac{(ae^{ax}h)^2}{2!} + ...}{h}$$

$$= \lim_{h \to 0} \left(ae^{ax} + \frac{ae^{ax}h}{2!} + ...\right)$$

$$= ae^{ax}$$

So, the derivative of $e^{ax}$ with respect to $x$ using first principles is $ae^{ax}$.