Derivatives of inverse exponential functions

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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:

\[ \frac {d}{dx} 7^x \begin{prompt} = \answer {7^x \ln (7)} \end{prompt} \]

2025-01-06 19:51:29