Introduction to differentiation: Derivatives of exponential functions and logarithms
Derivatives of exponential functions and logarithms
The exponential function and logarithm deserve special names because of the following special features.
Derivatives of the natural exponential function and natural logarithm
The derivative of the natural exponential function #\exp(x)# is #\exp(x)#.
The derivative of the natural logarithm #\ln(x)# is #\dfrac{1}{x}#.
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}#.
Derivatives of exponential and logarithmic functions
The derivative of the exponential function #f(t)=a^t# is equal to #f'(t)=\ln(a)\cdot a^t#.
The derivative of the logarithmic function #f(x) = \log_a(x)# is equal to #f'(x)=\frac{1}{x\cdot \ln(a)}#.
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:\[\begin{array}{rcl} \frac{\dd}{\dd x} \log_a(x) &=&\frac{\dd}{\dd x}\frac{\ln(x)}{\ln(a)} \\&&\phantom{xx}\color{blue}{\log_a(x)=\frac{\ln(x)}{\ln(a)}}\\&=&\frac{1}{\ln(a)}\cdot\frac{\dd}{\dd x}{\ln(x)}\\&&\phantom{xx}\color{blue}{\frac{\dd}{\dd x}\left(c\cdot f(x)\right)=c\cdot f'(x)}\\&=&\frac{1}{\ln(a)}\cdot\frac{1}{x}\\&&\phantom{xx}\color{blue}{\frac{\dd}{\dd x}\ln(x)=\frac{1}{x}}\\&=&\frac{1}{\ln(a)\cdot x}\\&&\phantom{xx}\color{blue}{\text{simplified}}\end{array}\]
According to the theory, the derivative of the exponential function #f(x)=a^{x}# is equal to #f'(x)=\ln(a) \cdot a^{x}#.
Substituting #a=3# yields #f'(x)=\ln \left(3\right)\cdot {3}^{x}#.
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