The natural logarithm differentiates to the reciprocal of its argument.
\(\ln x\) is the inverse of \(e^{x}\), and inverting a function reciprocates its slope. Since \(e^{x}\) has slope \(e^{x} = y\) at height \(y\), its inverse has slope \(\frac{1}{y}\) — at the point \(x\), that is \(\frac{1}{x}\).
This is the one derivative that fills the gap left by the power rule. Every power \(x^{n}\) integrates to another power except \(n = -1\); \(\ln x\) is what \(\frac{1}{x}\) integrates to.
Composed with something else, the pattern is \(\frac{u'}{u}\): derivative of the inside over the inside. That form is worth recognising on sight, because it appears everywhere in integration too.
Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.
Straight from the rule.
f rises without bound but ever more slowly, and f′ falls away toward zero. Near the origin the roles reverse: f dives toward \(-\infty\) while f′ climbs, both consequences of the same vertical tangent.
Defined only for \(x > 0\), where the derivative is positive and shrinking.
Open ln(x) in the calculatorChain rule: inside over inside.
The \(5\) cancels.
The curve is lifted by \(\ln 5 \approx 1.61\), yet f′ is exactly the \(\tfrac{1}{x}\) of the previous graph: a vertical shift changes no slope.
Unsurprising once you notice \(\ln 5x = \ln 5 + \ln x\): the two functions differ by a constant, so they have the same slope.
Open ln(5x) in the calculatorInside over inside, with \(u = x^{2} + 1\).
Tidy.
A symmetric bowl with its floor at the origin, flanked by an odd derivative that peaks at \(x = \pm 1\) and decays after.
The argument is never zero, so this derivative is defined for all real \(x\).
Open ln(x^2+1) in the calculatorProduct rule with \(f = x\), \(g = \ln x\).
The second term simplifies to \(1\).
f dips to a minimum at \(x = \tfrac{1}{e} \approx 0.368\), and f′ crosses zero at precisely that point.
This is why \(\int \ln x\,dx = x\ln x - x\).
Open x ln(x) in the calculatorInside over inside, with \(u = \sin x\).
Which is the cotangent.
Arches between the zeros of \(\sin x\), each plunging toward \(-\infty\) at both ends, with f′ running from \(+\infty\) to \(-\infty\) across every one.
A tidy example of the \(\frac{u'}{u}\) shape producing a familiar function.
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