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 $\ddx e^x = e^x$ , to deduce the derivative of $\ln (x)$ . We will use implicit differentiation to exploit this relationship computationally.

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

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\\ \ddx e^y &= \ddx x &\text{Differentiate both sides.}\\ e^y \dd[y]{x} &= 1 &\text{Implicit differentiation.}\\ \dd[y]{x} &= \frac{1}{e^y} = \answer[given]{\frac{1}{x}}&\text{Solve for $\dd[y]{x}$.} \end{align*}$

Since $y=\ln (x)$ , $\ddx \ln (x) = \answer [given]{\frac {1}{x}}$ .

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

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 $\ddx \log _b(x) = \frac {1}{x\ln (b)}.$

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

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

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

The derivative of an exponential function $\ddx a^x = a^x\cdot \ln (a).$

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

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