We previously saw that the derivative of $e^x$ was $e^x$. But what about other bases? To compute the derivative of $a^x$, we use the chain rule: \begin{eqnarray*} a = & e^{\ln(a)} & \hbox{ by the definition of $\ln(a)$} \cr \cr a^x = & (e^{\ln(a)})^x = e^{x \ln(a)} & \hbox{ by the laws of exponents} \cr\cr \frac{d (a^x)}{dx} = & \; e^{x \ln(a)} \, \frac{d}{dx}(x \ln(a)) & \hbox{ by the chain rule} \cr = & e^{x \ln(a)} \, \ln(a) & \cr = & a^x \, \ln(a). & \end{eqnarray*} The derivative of any exponential function $f(x)=a^x$ is proportional to $a^x$, and the proportionality constant is the natural log of $a$! There's no avoiding natural logs; they come up (well, naturally) whenever you look at rates of change of exponentials.
Next we turn to derivatives of logs. We defined log functions as inverses of exponentials: \begin{eqnarray*} y = \ln(x) & \Longleftrightarrow & x = e^y \cr y = \log_a(x) & \Longleftrightarrow & x = a^y. \end{eqnarray*} Since we know how to differentiate exponentials, we can use a technique called "implicit differentiation", which is based on the chain rule, to find the derivatives of $\ln(x)$ and $\log_a(x)$. The videos below walk us through this process.
The end results are:
| $$\frac{d}{dx} \ln(x) = \frac{1}{x}, \qquad \frac{d}{dx}\log_a(x) = \frac{1}{x \ln(a)}.$$ |
| The derivative of
$\ln(x)$: \begin{eqnarray*} y & = & \ln(x) \cr x & = & e^y \cr 1 & = & \frac{d }{dx}\left(e^y\right) \cr 1 & = & e^y \frac{dy}{dx} \cr \frac{dy}{dx} & = & \frac{1}{e^y} \cr \frac{dy}{dx} & = & \frac{1}{x} \end{eqnarray*}since $e^y=x$ (see above). |
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| The derivative of
$\log_a(x)$: \begin{eqnarray*} y & = & \log_a(x) \cr x & = & a^y \cr 1 & = & \frac{d}{dx} \left( a^y\right)\cr 1 & = & a^y \ln(a) \frac{dy}{dx} \cr \frac{dy}{dx} & = & \frac{1}{a^y \ln(a)} \cr \frac{dy}{dx} & = & \frac{1}{x \ln(a)}. \end{eqnarray*} since $a^y=x$ (see above). |
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