Derivative of the Natural Logarithm

The most elegant route goes via the inverse function. We have e^(ln x) = x for all x > 0. Differentiating both sides, the chain rule gives e^(ln x)·(ln x)′ on the left and 1 on the right. Since e^(ln x) = x, this reads x·(ln x)′ = 1, so (ln x)′ = 1/x. Intuitively: the ln curve is the exponential curve mirrored at the diagonal; where the exponential becomes steep, its inverse becomes flat. That is why the slope of ln x keeps decreasing as x grows: at x = 1 it is 1, at x = 2 only 0.5, at x = 100 just 0.01. The logarithm grows without bound, but ever more slowly.

Use the chain rule: outer derivative 1/u, inner derivative u′, together (ln u(x))′ = u′(x)/u(x). For ln(x² + 1): u = x² + 1, u′ = 2x, so the derivative is 2x/(x² + 1); at x = 1 this gives 2/2 = 1. The quotient u′/u is so important that it has its own name: logarithmic derivative. Read backwards it is an integration tool: if the numerator of a fraction is exactly the derivative of the denominator, the antiderivative is ln|denominator|, e.g. ∫2x/(x² + 1) dx = ln(x² + 1) + C. Common mistake: writing just 1/(x² + 1) and omitting the inner derivative 2x.

The natural logarithm is defined only for positive numbers, because eˣ takes only positive values; ln x answers the question "e to what power gives x?". So the derivative 1/x as the derivative of ln x also exists only for x > 0. When integrating, the perspective flips: 1/x is defined for negative x as well, and there ln(−x) is an antiderivative, since the chain rule gives (ln(−x))′ = (−1)/(−x) = 1/x. Both cases are combined in ln|x|: ∫1/x dx = ln|x| + C, valid on every interval that does not contain 0. Even so, you must never integrate across the pole at x = 0.

Via the change of base: log_b x = ln x/ln b, where ln b is a constant. Differentiating the constant multiple gives (log_b x)′ = 1/(x·ln b). For the base-10 logarithm this means (lg x)′ = 1/(x·ln 10) ≈ 1/(2.303·x), for the base-2 logarithm (log₂ x)′ = 1/(x·ln 2) ≈ 1/(0.693·x). Only the natural logarithm has the clean derivative 1/x without a prefactor, which makes it the standard choice in calculus. The typical mistake is writing (lg x)′ = 1/x; the result is then too large by the factor ln 10 ≈ 2.3. Rule of thumb: first rewrite a foreign base in ln, then differentiate.

Not 1/x, that is the derivative! You find the antiderivative with integration by parts and the trick of reading ln x as the product 1·ln x: choose u = ln x and v′ = 1, so u′ = 1/x and v = x. Then ∫ln x dx = x·ln x − ∫x·(1/x) dx = x·ln x − ∫1 dx = x·ln x − x + C. Check by differentiating: (x·ln x − x)′ = ln x + x·(1/x) − 1 = ln x ✓. This task is an exam classic precisely because the confusion with 1/x is so tempting. Remember the directions: differentiating turns ln x into the fraction 1/x, integrating turns it into x·ln x − x.