Derivative and direction of change of ln

The derivative of the natural logarithm

The function ln is differentiable on the interval ]0; +∞[ and its derivative is given by:

ln'(x) = 1/x

This formula is essential: it allows us to differentiate any expression containing a ln and, conversely, to recognise an antiderivative in the form 1/x.

Direction of variation

For all x > 0, we have 1/x > 0. The derivative of ln is therefore strictly positive over its entire domain of definition: the function ln is strictly increasing on ]0; +∞[.

This means that for all real numbers a > 0 and b > 0:

a < b ⇔ ln(a) < ln(b)

a = b ⇔ ln(a) = ln(b)

This equivalence is widely used to solve inequalities: one can apply ln to both sides of an inequality involving strictly positive real numbers without changing the direction of the inequality.

Derivative of a composite function

If u is a differentiable function that is strictly positive on an interval I, then the function f defined by f(x) = ln(u(x)) is differentiable on I and:

f'(x) = u'(x) / u(x)

Example

Let f(x) = ln(x² + 1). Here, u(x) = x² + 1 is always strictly positive, and u'(x) = 2x. Therefore:

f'(x) = 2x / (x² + 1)

The sign of f'(x) is that of 2x (since x² + 1 is always > 0): f is therefore decreasing on ]–∞; 0] and increasing on [0; +∞[.

Pitfall to avoid

Do not confuse ln'(x) = 1/x with the derivative of ln(u(x)), which is u'(x)/u(x). Forgetting to multiply by u'(x) is the most common mistake when differentiating a composite function involving ln.