First we determine the derivative of #\exp# in #a#. The difference quotient can be rewritten as follows:\[\frac{\exp(a+h)-\exp(a)}{h} = \frac{{\e}^{a+h}- {\e}^a}{h} = {\e}^a\frac{{\e}^{h}-1}{h}\tiny.\]

It is known that #\lim_{h\to 0}\frac{{\e}^h-1}{h} = 1#. It follows that:\[\begin{array}{rcl}{\exp}'(a)&=& \lim_{h\to 0}\frac{\exp(a+h)-\exp(a)}{h} \\ &=& \lim_{h\to 0}{\e}^a\frac{{\e}^{h}-1}{h}\\ &=& {\e}^a\cdot\lim_{h\to 0}\frac{{\e}^{h}-1}{h} \\ &=& {\e}^a=\exp(a)\end{array}\]Thus, the derivative function of #\exp# is #\exp# itself.

Later we will prove that the derivative of #\ln(x)# equals #\dfrac{1}{x}#.

The first statement follows from:\[\begin{array}{rcl} \frac{\dd}{\dd t}\left(a^t\right) &=& \lim_{h\to 0}\frac{a^{t+h}-a^t}{h}\\&&\phantom{xx}\color{blue}{\text{definition derivative}}\\ &=&\lim_{h\to 0}\frac{\e^{\ln(a)\cdot(t+h)}-\e^{\ln(a)\cdot t}}{h}\\&&\phantom{xx}\color{blue}{a^x=\e^{\ln(a)\cdot x}}\\ &=&\lim_{h\to 0}\left(\ln(a)\cdot\frac{\e^{\ln(a)\cdot t+\ln(a)\cdot h}-\e^{\ln(a)\cdot t}}{\ln(a)\cdot h}\right)\\&&\phantom{xx}\color{blue}{\text{numerator and denominator multiplied by }\ln(a)}\\ &=&\ln(a)\cdot \lim_{h\to 0}\frac{\e^{\ln(a)\cdot t+\ln(a)\cdot h}-\e^{\ln(a)\cdot t}}{\ln(a)\cdot h}\\&&\phantom{xx}\color{blue}{\lim_{h\to0}\left(\ln(a)\cdot g(h)\right)=\ln(a)\cdot\lim_{h\to0}g(h)}\\&=&\ln(a)\cdot \lim_{k\to 0}\frac{\e^{\ln(a)\cdot t+k}-\e^{\ln(a)\cdot t}}{k}\\&&\phantom{xx}\color{blue}{k=\ln(a)\cdot h}\\&=&\ln(a)\cdot{\exp}'(\ln(a)\cdot t)\\&&\phantom{xx}\color{blue}{\text{definition derivative}}\\&=&\ln(a)\cdot{\exp}(\ln(a)\cdot t)\\&&\phantom{xx}\color{blue}{\exp'=\exp}\\&=&\ln(a)\cdot{\e}^{\ln(a)\cdot t}\\&&\phantom{xx}\color{blue}{\text{definition }\exp}\\&=&\ln(a)\cdot{a}^{t}\\&&\phantom{xx}\color{blue}{\e^{\ln(a)\cdot t}=a^t}\end{array}\]The second statement follows from:\[\begin{array}{rcl} \frac{\dd}{\dd x} \log_a(x) &=&\frac{\dd}{\dd x}\frac{\ln(x)}{\ln(a)} \\&&\phantom{xx}\color{blue}{\log_a(x)=\frac{\ln(x)}{\ln(a)}}\\&=&\frac{1}{\ln(a)}\cdot\frac{\dd}{\dd x}{\ln(x)}\\&&\phantom{xx}\color{blue}{\frac{\dd}{\dd x}\left(c\cdot f(x)\right)=c\cdot f'(x)}\\&=&\frac{1}{\ln(a)}\cdot\frac{1}{x}\\&&\phantom{xx}\color{blue}{\frac{\dd}{\dd x}\ln(x)=\frac{1}{x}}\\&=&\frac{1}{\ln(a)\cdot x}\\&&\phantom{xx}\color{blue}{\text{simplified}}\end{array}\]