Derivatives of inverse exponential functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We derive the derivatives of inverse exponential functions using implicit differentiation.

Geometrically, there is a close relationship between the plots of $e^x$ and $\ln (x)$ , they are reflections of each other over the line $y=x$ :

One may suspect that we can use the fact that $\frac {d}{dx} e^x = e^x$ , to deduce the derivative of $\ln (x)$ . We will use implicit differentiation to exploit this relationship computationally.

The Derivative of the Natural Logrithm $\frac {d}{dx} \ln (x) = \frac {1}{x}.$

Recall $\ln (x) = y \qquad \text {exactly when}\qquad e^y = x\qquad \text {and}\qquad x>\answer [given]{0}.$ Hence $\begin{align*} e^y &= x\\ \frac{d}{dx} e^y &= \frac{d}{dx} x &\text{Differentiate both sides.}\\ e^y \frac{dy}{dx} &= 1 &\text{Implicit differentiation.}\\ \frac{dy}{dx} &= \frac{1}{e^y} = \answer[given]{\frac{1}{x}}&\text{Solve for $\frac{dy}{dx}$.} \end{align*}$ Since $y=\ln (x)$ , $\frac {d}{dx} \ln (x) = \answer [given]{\frac {1}{x}}$ .

Compute: $\frac {d}{dx} \left (-\ln (\cos (x))\right ) \begin {prompt} = \answer {\tan (x)} \end {prompt}$

From the derivative of the natural logarithm, we can deduce another fact:

The derivative of any logarithm Let $b$ be a positive real number. Then $\frac {d}{dx} \log _b(x) = \frac {1}{x\ln (b)}.$

Here we need to remember that $\log _b(x) = \frac {\ln (x)}{\answer [given]{\ln (b)}}.$ So we may write $\begin{align*} \frac{d}{dx} \log_b(x) &= \frac{d}{dx} \frac{\ln(x)}{\answer[given]{\ln(b)}}\\ &= \answer[given]{\frac{1}{x\ln(b)}}. \end{align*}$

Compute: $\frac {d}{dx} \log _7(x) \begin {prompt} = \answer {1/(x \ln (7))} \end {prompt}$

We can also compute the derivative of an arbitrary exponential function.

The derivative of an exponential function $\frac {d}{dx} a^x = a^x\cdot \ln (a).$

Here we need to be slightly sneaky. Note $a^x = e^{\ln (a^x)} = e^{x\ln (a)}.$ So we may write $\begin{align*} \frac{d}{dx} a^x &= \frac{d}{dx} e^{x\ln(a)}\\ &= e^{x\ln(a)} \cdot \answer[given]{\ln(a)}\\ &= \answer[given]{a^x\cdot \ln(a)}. \end{align*}$

Compute: $\frac {d}{dx} 7^x \begin {prompt} = \answer {7^x \ln (7)} \end {prompt}$