Differentiate e^(ax)eax 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)