Derivative of ln(x) is 1/x

Q: What is the derivative of ln(x)?

The derivative of ln(x) is 1/x, for x > 0 only.

Answer

\frac{1}{x}

Derivative of ln(x)

Rule Used: Logarithmic Rule
Steps: Recognize the natural logarithm $\log(x)$ , written $\ln(x)$ on this site. Its derivative is the reciprocal because ln is the inverse of e^x and the derivative of e^x is itself.
Apply the rule: $\frac{d}{dx} \log(x)$ = 1/x.
Final answer: $x^{-1}$

The derivative of ln(x) is 1/x, defined only for x > 0.

Instant step-by-step solution for ln(x)

Derivative LogarithmicIntermediate

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Derivative of ln(x) = x^(-1) Solve step-by-step for free: https://derivativecalculatorai.com/derivative-of-ln-x

Derivative

ln(x) → x^(-1)engine: x^(-1)

Graph of the function and its derivative

The solid curve is f. The dashed curve is f′. Vertical gaps are asymptotes or jumps, not extra zeros.

Why the derivative has this form

The natural logarithm has no values for x <= 0, so neither does its derivative. Writing 1/x without that restriction is shorthand; the full statement is d/dx ln(x) = 1/x on the domain x > 0. The related function ln|x| is defined on both sides of zero and has derivative 1/x for every nonzero x, which is where the two formulas are often mixed up.

The cleanest proof treats ln(x) as the inverse of e^y. Set y = ln(x), so x = e^y. Differentiating both sides with respect to x gives 1 = e^y·dy/dx, and since e^y = x, solving for dy/dx yields 1/x. The derivative of e^y does all the work, and it is why the natural logarithm, not log base 10, differentiates so simply.

The graph agrees. Near x = 0 the logarithm dives steeply toward negative infinity and 1/x is large; as x grows the curve flattens and 1/x tends to 0. The slope is always positive, so ln(x) is increasing on its whole domain, but it grows more and more slowly. For log base 10 the extra factor 1/ln(10) appears, roughly 0.434/x.

Inverse-function proof from x = e^y

Set y = ln(x). By the definition of the natural log, x = e^y.
Differentiate both sides with respect to x: 1 = e^y·dy/dx.
Substitute e^y = x back into that equation: 1 = x·dy/dx.
Solve for dy/dx = 1/x, valid where x > 0.
State the domain explicitly: ln(x) is undefined for x <= 0, so 1/x is a derivative there only on the positive half-line.
Contrast with log base 10: d/dx log10(x) = 1/(x ln(10)). The change-of-base factor survives differentiation.
Contrast with ln|x|: on the negative half-line its derivative is also 1/x, but the function itself is ln(-x) there.

Worked example

Slope of ln(x) at x = e and at x = 1

At x = e the function value is ln(e) = 1 and the derivative is 1/e, about 0.368.
The tangent line at that point is y - 1 = (1/e)(x - e), which simplifies to y = x/e.
At x = 1 the value is 0 and the derivative is 1, so the tangent there is y = x - 1.
A secant from 1 to 1.001 gives ln(1.001)/0.001, about 0.9995, confirming the slope near 1 is 1.

Answer: 1/e

Common mistakes

Writing 1/|x| for ln(x)

ln(x) is only defined for positive x, so 1/x is already positive on its whole domain. The absolute value belongs to the derivative of ln|x|, a different function defined for negative x as well.

Forgetting the ln(10) factor for log(x)

If log means base 10, the derivative is 1/(x ln(10)), not 1/x. Only the natural logarithm, base e, differentiates to 1/x without an extra constant.

Evaluating at x = 0 or a negative x

ln(0) and ln(-3) are undefined, so there is no slope at those points. The expression 1/x may produce a number, but a derivative cannot exist where the function has no value.

Derivative of e^x — e^x. The inverse function used in the proof above.
Logarithmic derivatives — The rule page lists this example and the others in the same family.
Exponential derivatives — The rule page lists this example and the others in the same family.
Derivative of ln(2x) — Same family of differentiation, different function.
Derivative of log(x) — Same family of differentiation, different function.