\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.6in,top=0.6in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Differentiating Exponential Functions} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, 2015} \rfoot{\small Last revised: October 29, 2015} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} Recall that when we compose a function with its inverse, we obtain the identity function. \ \begin{equation*} f\left( f^{-1}\left( x\right) \right) =x\text{ \ for all }x \end{equation*}% Recall the chain rule: \ for differentiable functions $f$ and $g$, \begin{equation*} \left[ f\left( g\left( x\right) \right) \right] ^{\prime }=f^{\prime }\left( g\left( x\right) \right) \cdot g^{\prime }\left( x\right) \end{equation*}% These two theorems are what we need to differentiate exponential functions. \ We will start with $g\left( x\right) =e^{x}$. \ We already know that $% g\left( x\right) =e^{x}$ and $f\left( x\right) =\ln x$ are inverses of each other, i.e.% \begin{equation*} \ln \left( e^{x}\right) =x \end{equation*}% We differentiate both sides of this equation. \ For the left-hand side, we use the chain rule.% \begin{eqnarray*} \left[ \ln \left( e^{x}\right) \right] ^{\prime } &=&\left( x\right) ^{\prime } \\ \dfrac{1}{e^{x}}\cdot \left[ \left( e^{x}\right) ^{\prime }\right] &=&1% \text{ \ \ \ \ \ \ \ \ we solve for }\left( e^{x}\right) ^{\prime } \\ \left( e^{x}\right) ^{\prime } &=&e^{x} \end{eqnarray*}% So, we have proved that $\left( e^{x}\right) ^{\prime }=e^{x}$. \ This is of course a very surprising and interesting property that we have only seen thus far with the constant zero function. \ It also quickly follows that the higher-order derivates are the same, i.e. \ if $f\left( x\right) =e^{x}$, then $f^{\prime }\left( x\right) =e^{x},$ \ \ \ $f^{\prime \prime }\left( x\right) =e^{x}$, \ \ $f^{\prime \prime \prime }\left( x\right) =e^{x}$, \ \ \ $f^{\left( 4\right) }\left( x\right) =e^{x}$ \ and so on. \ It is also clear that the antiderivative of $e^{x}$ is also the function itself, up to a constant added: \begin{equation*} \int e^{x}dx=e^{x}+C \end{equation*} Example: \ Differentiate $h\left( x\right) =e^{7x}$. Solution: We will use the chain rule. \ We define the outer function, $% f\left( x\right) =e^{x}$ and the inner function, $g\left( x\right) =7x$. \ \ Then% \begin{eqnarray*} \left[ f\left( g\left( x\right) \right) \right] ^{\prime } &=&f^{\prime }\left( g\left( x\right) \right) \cdot g^{\prime }\left( x\right) \text{ becomes} \\ \left[ e^{7x}\right] ^{\prime } &=&e^{7x}\cdot 7=7e^{x} \end{eqnarray*} Example: \ Differentiate $h\left( x\right) =\dfrac{1}{e^{4x}}$. Solution: \ We can re-write $\dfrac{1}{e^{3x}}$ as $e^{-3x}$ and apply the chain rule as before. \ We define the outer function, $f\left( x\right) =e^{x}$ and the inner function, $g\left( x\right) =-3x$. \ \ Then% \begin{eqnarray*} \left[ f\left( g\left( x\right) \right) \right] ^{\prime } &=&f^{\prime }\left( g\left( x\right) \right) \cdot g^{\prime }\left( x\right) \text{ becomes} \\ \left[ e^{-3x}\right] ^{\prime } &=&e^{-3x}\cdot \left( -3\right) =\dfrac{-3% }{e^{3x}} \end{eqnarray*} What about other exponential functions such as $f\left( x\right) =5^{x}$? \ We can also differentiate those using the chain rule. \ Recall first that \begin{equation*} 5^{x}=e^{\ln \left( 5^{x}\right) } \end{equation*}% This is true because $e^{x}$ and $\ln x$ are inverses of each other, and so $% e^{\ln A}=A$ \ \ for all positive $A$. \ We substitute $A=5^{x}$ and obtain \begin{equation*} e^{\ln \left( 5^{x}\right) }=5^{x} \end{equation*}% Why does this help? \ If we re-write $5^{x}$ as $e^{\ln \left( 5^{x}\right) } $, then we can use properties of logarithms, the chain rule, and that $% \left( e^{x}\right) ^{\prime }=e^{x}$ to differentiate $5^{x}$. \begin{equation*} f\left( x\right) =5^{x}=e^{\ln \left( 5^{x}\right) }=e^{x\ln 5}=e^{\left( \ln 5\right) x} \end{equation*}% Notice that $\ln 5$ is just a constant so the function $e^{\left( \ln 5\right) x}$ is very similar to a function like $e^{7x}$ and can be easily differentiated by the chain rule. \ We define the outer function, $f\left( x\right) =e^{x}$ and the inner function, $g\left( x\right) =\left( \ln 5\right) x$. \ \ Then% \begin{eqnarray*} \left[ f\left( g\left( x\right) \right) \right] ^{\prime } &=&f^{\prime }\left( g\left( x\right) \right) \cdot g^{\prime }\left( x\right) \text{ becomes} \\ \left[ e^{\left( \ln 5\right) x}\right] ^{\prime } &=&e^{\left( \ln 5\right) x}\cdot \left( \ln 5\right) \end{eqnarray*}% The first factor, $e^{\left( \ln 5\right) x}$ is still the same $5^{x}$, and so we have that \begin{equation*} \left( 5^{x}\right) ^{\prime }=5^{x}\cdot \ln 5 \end{equation*} \end{document}