Derivatives of the exponential and the logarithm

This is the story of the function that does not change under differentiation and the function whose derivative is $\dfrac{1}{x}$. We look at $y = e^x$ and $y = \ln x$.

The slope equals the height

At every point of $e^x$ the slope of the tangent equals the height of the point¹.

The only functions with that property are those of the form $Ce^x$, and the number $e$ is fixed by it.

The same does not happen for $2^x$. There $\left( 2^x \right)' = 2^x \ln 2$, off by the factor $\ln 2 \approx 0.693$. Only with base $e$ does the extra coefficient disappear.

The derivative of the logarithm

The derivative of $\ln x$ is $\dfrac{1}{x}$².

It flattens toward the right. That is why $\ln x$ increases without bound while its rate of increase falls away.

The relation between inverses

The two are inverse to each other, and their graphs are symmetric about $y = x$. Since the derivative of an inverse is $\dfrac{dx}{dy} = \dfrac{1}{dy/dx}$, if the slope of $y = e^x$ is $y$ itself then the slope of $x = \ln y$ is $\dfrac{1}{y}$. Each follows from the other.

Filling a hole in integration

Integrating $\dfrac{1}{x}$ returns $\ln x$. The integral of $x^n$ is $\dfrac{x^{n+1}}{n+1}$, which fails at $n = -1$ because the denominator becomes $0$. The logarithm is what fills that hole.

When the base is not $e$

The first follows from rewriting $a^x = e^{x \ln a}$ and applying the chain rule, the second from $\log_a x = \dfrac{\ln x}{\ln a}$. Whatever the base, converting to natural logarithms reduces everything to the same two formulas.

The steeply rising curve on the graph is $y = e^x$, the gently rising one is $y = \ln x$, the hyperbola is $y = \dfrac{1}{x}$, and the large dots are $(0, 1)$, $(1, 0)$ and $(1, 1)$.

1. Exponential function, Wikipedia
2. Natural logarithm, Wikipedia