Natural Logarithm: Definition, Derivative & Rules

#Calculus
TL;DR
The natural logarithm, written ln⁡x\ln x, is the logarithm with base e≈2.71828e \approx 2.71828, so ln⁡x=log⁡ex\ln x = \log_e x. In calculus it is defined as the area under the curve y=1ty = \tfrac{1}{t} from t=1t=1 to t=xt=x, which is why ln⁡1=0\ln 1 = 0 and ln⁡e=1\ln e = 1. Its derivative is the cleanest in all of calculus, ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x}, and it is the antiderivative that fills the gap left by the power rule: ∫1x,dx=ln⁡∣x∣+C\int \frac{1}{x},dx = \ln|x| + C.
BT
Bhanzu TeamLast updated on September 23, 202613 min read

What Is The Natural Logarithm?

The natural logarithm of a positive number xx is the power to which ee must be raised to give xx. In symbols, ln⁡x=log⁡ex\ln x = \log_e x, where e≈2.71828e \approx 2.71828 is Euler's number. So ln⁡x=y\ln x = y means exactly ey=xe^{y} = x.

That makes the natural logarithm the inverse of the exponential function exe^{x}. Feeding a number into exe^{x} and then into ln⁡\ln returns the original number, and the two undo each other:

ln⁡(ex)=x,eln⁡x=x(x>0)\ln(e^{x}) = x, \qquad e^{\ln x} = x \quad (x > 0)

Three values anchor the whole function:

  • ln⁡1=0\ln 1 = 0, because e0=1e^{0} = 1.

  • ln⁡e=1\ln e = 1, because e1=ee^{1} = e.

  • ln⁡x\ln x is only defined for x>0x > 0; you cannot take the logarithm of zero or a negative number in the real numbers.

The algebra of ln⁡\ln as "the log with base ee" is covered on the companion page for logarithmic functions. This page takes the calculus view: where ln⁡\ln comes from as an area, how it differentiates and integrates, and why it earns the word "natural."

How Is The Natural Logarithm Defined As An Integral?

Calculus gives the natural logarithm a definition that needs no mention of exponents at all. For x>0x > 0:

ln⁡x=∫1x1t,dt\ln x = \int_{1}^{x} \frac{1}{t},dt

Read geometrically, ln⁡x\ln x is the signed area under the curve y=1ty = \tfrac{1}{t} between t=1t = 1 and t=xt = x. This one picture explains the anchor values instantly.

  • Why ln⁡1=0\ln 1 = 0. The area from 11 to 11 has zero width, so ln⁡1=∫111t,dt=0\ln 1 = \int_{1}^{1}\frac{1}{t},dt = 0.

  • Why ln⁡e=1\ln e = 1. The number ee is defined as the point where the running area first reaches exactly 11: ∫1e1t,dt=1\int_{1}^{e}\frac{1}{t},dt = 1. That is what makes e≈2.71828e \approx 2.71828 special rather than arbitrary.

  • Why ln⁡x<0\ln x < 0 for 0<x<10 < x < 1. Integrating "backwards" from 11 down to a smaller number flips the sign, so the logarithm of a number below 11 is negative.

What Is The Derivative Of The Natural Logarithm?

The integral definition pays off immediately. By the Fundamental Theorem of Calculus, differentiating an "area so far" function returns the curve you were measuring under. Since ln⁡x\ln x is the area under 1t\tfrac{1}{t}, its derivative is 1x\tfrac{1}{x}:

ddx[ln⁡x]=1x,x>0\frac{d}{dx}[\ln x] = \frac{1}{x}, \qquad x > 0

This is the simplest derivative of any standard function, and it is the geometric heart of the topic: the slope of the tangent to the ln⁡\ln curve at any point xx is exactly 1x\tfrac{1}{x}. Near x=1x=1 the graph climbs with slope 11; far to the right it flattens, because 1x\tfrac{1}{x} shrinks toward 00.

For negative inputs the absolute-value version holds for every x≠0x \neq 0:

ddx[ln⁡∣x∣]=1x\frac{d}{dx}\big[\ln|x|\big] = \frac{1}{x}

The check is quick. For x>0x > 0, ln⁡∣x∣=ln⁡x\ln|x| = \ln x, whose derivative is 1x\tfrac{1}{x}. For x<0x < 0, ln⁡∣x∣=ln⁡(−x)\ln|x| = \ln(-x), and the chain rule gives 1−x⋅(−1)=1x\frac{1}{-x}\cdot(-1) = \tfrac{1}{x}. Both cases land on 1x\tfrac{1}{x}.

Most real problems wrap the logarithm around another function. The chain rule then gives the single most-used differentiation formula in this topic:

ddx[ln⁡(g(x))]=g′(x)g(x)\frac{d}{dx}\big[\ln(g(x))\big] = \frac{g'(x)}{g(x)}

In words: the derivative of the log of something is the derivative of the inside, divided by the inside. Forgetting that inner derivative g′(x)g'(x) is the classic slip, and it has its own entry in the mistakes section below.

How Do You Integrate 1/x With The Natural Logarithm?

The power rule ∫xn,dx=xn+1n+1+C\int x^{n},dx = \frac{x^{n+1}}{n+1} + C breaks at n=−1n = -1, because it would divide by zero. The natural logarithm is exactly the function that fills that hole:

∫1x,dx=ln⁡∣x∣+C\int \frac{1}{x},dx = \ln|x| + C

The absolute value matters. The integrand 1x\tfrac{1}{x} is defined for negative xx too, so the antiderivative must be as well, and ln⁡∣x∣\ln|x| covers both sides of zero. Differentiating ln⁡∣x∣+C\ln|x| + C returns 1x\tfrac{1}{x}, which confirms the formula.

Wrapped around an inner function, this becomes the pattern that solves a huge share of integrals, any fraction whose numerator is the derivative of its denominator:

∫g′(x)g(x),dx=ln⁡∣g(x)∣+C\int \frac{g'(x)}{g(x)},dx = \ln|g(x)| + C

What Are The Log Laws Of The Natural Logarithm?

Because ln⁡\ln is a logarithm, it turns products into sums, quotients into differences, and powers into multiples. For a,b>0a, b > 0 and any real rr:

Table: The natural-logarithm laws, with the everyday meaning of each.

Law

Statement

What it does

Product

ln⁡(ab)=ln⁡a+ln⁡b\ln(ab) = \ln a + \ln b

Multiplication becomes addition

Quotient

ln⁡!(ab)=ln⁡a−ln⁡b\ln!\left(\dfrac{a}{b}\right) = \ln a - \ln b

Division becomes subtraction

Power

ln⁡(ar)=r,ln⁡a\ln(a^{r}) = r,\ln a

An exponent comes out front as a factor

Identity

ln⁡1=0,ln⁡e=1\ln 1 = 0,\quad \ln e = 1

The two anchor values

These are not accidents of the base; the integral definition proves them. Take the product law. Define f(x)=ln⁡(ax)f(x) = \ln(ax) for a fixed constant a>0a > 0. By the chain rule, f′(x)=1ax⋅a=1xf'(x) = \frac{1}{ax}\cdot a = \frac{1}{x}, which is the same derivative as ln⁡x\ln x.

Two functions with the same derivative differ only by a constant, so ln⁡(ax)=ln⁡x+C\ln(ax) = \ln x + C. Setting x=1x = 1 gives ln⁡a=0+C\ln a = 0 + C, so C=ln⁡aC = \ln a. Therefore ln⁡(ax)=ln⁡x+ln⁡a\ln(ax) = \ln x + \ln a, and putting x=bx = b yields ln⁡(ab)=ln⁡a+ln⁡b\ln(ab) = \ln a + \ln b; the other laws follow the same way. For the full algebraic treatment, see logarithm rules and properties of logarithms.

How Do You Change The Base To The Natural Logarithm?

Any logarithm can be rewritten in terms of the natural logarithm using the change-of-base formula. For a base b>0b > 0, b≠1b \neq 1:

log⁡bx=ln⁡xln⁡b\log_{b} x = \frac{\ln x}{\ln b}

This is why a calculator with only ln⁡\ln and log⁡10\log_{10} keys can still find, say, log⁡210\log_{2} 10:

log⁡210=ln⁡10ln⁡2=2.3025850.693147=3.3219\log_{2} 10 = \frac{\ln 10}{\ln 2} = \frac{2.302585}{0.693147} = 3.3219

Change of base also explains the derivative of a general logarithm. Writing log⁡bx=ln⁡xln⁡b\log_{b} x = \frac{\ln x}{\ln b} and differentiating (with ln⁡b\ln b a constant) gives:

ddx[log⁡bx]=1x,ln⁡b\frac{d}{dx}\big[\log_{b} x\big] = \frac{1}{x,\ln b}

The extra ln⁡b\ln b in the denominator is the "tax" for using any base other than ee. Base ee is the only one where that factor equals 11 and the derivative is simply 1x\tfrac{1}{x}, which is a first hint at why it is called natural. Rewriting between exponential and log form is covered on the exponential form page.

How Do You Work With The Natural Logarithm? (Worked Examples)

Each example below is checked by differentiating the answer back to the starting expression, or by re-deriving the value.

Example 1: Differentiate y=ln⁡(3x2+1)y = \ln(3x^{2} + 1).

Use the chain rule ddx[ln⁡(g(x))]=g′(x)g(x)\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)} with g(x)=3x2+1g(x) = 3x^{2} + 1, so g′(x)=6xg'(x) = 6x:

dydx=6x3x2+1\frac{dy}{dx} = \frac{6x}{3x^{2} + 1}

Check: the inside derivative is 6x6x and the inside is 3x2+13x^{2}+1, matching the formula.

Final answer: dydx=6x3x2+1\dfrac{dy}{dx} = \dfrac{6x}{3x^{2} + 1}.

Example 2: Differentiate y=ln⁡(sin⁡x)y = \ln(\sin x).

Here g(x)=sin⁡xg(x) = \sin x and g′(x)=cos⁡xg'(x) = \cos x:

dydx=cos⁡xsin⁡x=cot⁡x\frac{dy}{dx} = \frac{\cos x}{\sin x} = \cot x

Final answer: dydx=cot⁡x\dfrac{dy}{dx} = \cot x. (This is one of the ways the natural logarithm connects to trigonometry.)

Example 3: Evaluate ∫2xx2+1,dx\displaystyle\int \frac{2x}{x^{2} + 1},dx.

The numerator 2x2x is exactly the derivative of the denominator x2+1x^{2} + 1, so this fits ∫g′(x)g(x),dx=ln⁡∣g(x)∣+C\int \frac{g'(x)}{g(x)},dx = \ln|g(x)| + C. Let u=x2+1u = x^{2} + 1, so du=2x,dxdu = 2x,dx:

∫2xx2+1,dx=∫1u,du=ln⁡∣u∣+C=ln⁡(x2+1)+C\int \frac{2x}{x^{2} + 1},dx = \int \frac{1}{u},du = \ln|u| + C = \ln(x^{2} + 1) + C

Because x2+1>0x^{2} + 1 > 0 for every real xx, the absolute-value bars are optional here.

Check: ddx[ln⁡(x2+1)+C]=2xx2+1\dfrac{d}{dx}\big[\ln(x^{2}+1) + C\big] = \dfrac{2x}{x^{2}+1}, the original integrand.

Final answer: ln⁡(x2+1)+C\ln(x^{2} + 1) + C.

Example 4: Evaluate the definite integral ∫1e1x,dx\displaystyle\int_{1}^{e} \frac{1}{x},dx.

∫1e1x,dx=[ln⁡∣x∣]1e=ln⁡e−ln⁡1=1−0=1\int_{1}^{e} \frac{1}{x},dx = \Big[\ln|x|\Big]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1

Final answer: 11. This is the definition of ee read as an area: the region under 1x\tfrac{1}{x} from 11 to ee has area exactly 11.

Example 5: Find xx if ln⁡x=2\ln x = 2, to four decimal places.

Rewrite in exponential form: ln⁡x=2\ln x = 2 means x=e2x = e^{2}.

x=e2=2.718282=7.3891x = e^{2} = 2.71828^{2} = 7.3891

Check: ln⁡7.3891≈2\ln 7.3891 \approx 2. Final answer: x=7.3891x = 7.3891.

A reference table collects the results used above.

Table: Core derivative and integral forms for the natural logarithm.

Expression

Result

Note

ddx[ln⁡x]\dfrac{d}{dx}[\ln x]

1x\dfrac{1}{x}

x>0x > 0

$\dfrac{d}{dx}[\ln

x

]$

ddx[ln⁡(g(x))]\dfrac{d}{dx}[\ln(g(x))]

g′(x)g(x)\dfrac{g'(x)}{g(x)}

Chain rule

∫1x,dx\displaystyle\int \dfrac{1}{x},dx

$\ln

x

∫g′(x)g(x),dx\displaystyle\int \dfrac{g'(x)}{g(x)},dx

$\ln

g(x)

For the wider families these belong to, see the derivative and integration formula pages.

Why Are e And The Natural Logarithm Called "Natural"?

The name is not marketing. Base ee is the one choice that removes every stray constant from calculus, and the reasons stack up.

  • The exponential is its own derivative. Only y=exy = e^{x} satisfies dydx=ex\frac{dy}{dx} = e^{x}; its rate of change equals its current value at every point. For any other base, ddx[bx]=bxln⁡b\frac{d}{dx}[b^{x}] = b^{x}\ln b carries an extra ln⁡b\ln b factor.

  • The derivative of ln⁡\ln is clean. As shown above, ddx[log⁡bx]=1xln⁡b\frac{d}{dx}[\log_{b} x] = \frac{1}{x\ln b}, and only b=eb = e makes that factor 11, leaving the tidy 1x\tfrac{1}{x}.

  • It is the mathematics of continuous growth. Anything that grows or decays at a rate proportional to its current size (money compounding, a population, a cooling body, a decaying isotope) is governed by exe^{x}, and reading the time or rate back out of it requires ln⁡\ln. The constant itself is the limit of continuous compounding, e=lim⁡n→∞(1+1n)ne = \displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}.

So ln⁡\ln is "natural" in the precise sense that it is the logarithm calculus would invent on its own, with no arbitrary base bolted on. It is the inverse of the one exponential whose growth rate is its own size. The link to exponential and inverse behaviour is developed on the inverse functions page.

Who Discovered The Natural Logarithm?

The natural logarithm arrived twice: first as a mysterious area under a curve, and only later as "the log with base ee." The area came first.

Two more figures completed the picture:

  • John Napier (1550–1617, Scotland) published the first logarithm tables in 1614, inventing logarithms as a tool to turn multiplication into addition for astronomers and navigators.

  • Leonhard Euler (1707–1783, Switzerland) named the constant ee, showed that e=lim⁡n→∞(1+1n)ne = \lim_{n\to\infty}(1+\tfrac{1}{n})^{n}, and made ln⁡\ln the inverse of exe^{x} the standard view. The base of the natural logarithm carries his initial to this day.

Where Is The Natural Logarithm Used In The Real World?

Any process with a rate proportional to its own size is governed by ee and read back with ln⁡\ln.

  • Radioactive decay and carbon dating. A sample decays as N=N0e−λtN = N_{0}e^{-\lambda t}, and the half-life is t1/2=ln⁡2λt_{1/2} = \dfrac{\ln 2}{\lambda}. The natural logarithm is what converts a decay rate into an age.

  • Continuous growth and finance. Continuously compounded money grows as A=PertA = Pe^{rt}; the time for an investment to double is ln⁡2r\dfrac{\ln 2}{r}, the same ln⁡2≈0.6931\ln 2 \approx 0.6931 that appears in half-life.

  • Information theory. Entropy measured in nats uses the natural logarithm, H=−∑p,ln⁡pH = -\sum p,\ln p, quantifying uncertainty in a signal or a message.

  • Chemistry and seismology. The pH scale and the Richter magnitude scale are logarithmic, compressing enormous ranges into readable numbers; the same logarithmic compression, in base ee, drives reaction-rate and cooling models.

  • Machine learning. Log-likelihood and cross-entropy loss functions are built on ln⁡\ln, because turning products of probabilities into sums of logarithms keeps the numbers stable and the calculus simple.

One function connects an isotope's age, a bank balance, and a message's information content, because all three change at a rate set by their own size.

What Are The Most Common Mistakes With The Natural Logarithm?

These four errors account for most lost marks on the natural logarithm, verified against Johns Hopkins and UConn logarithm notes, Britannica's natural-log entry, and the recurring r/learnmath and r/calculus questions behind them.

Splitting the log of a sum.

Where it slips in:

A student rewrites ln⁡(a+b)\ln(a + b) as ln⁡a+ln⁡b\ln a + \ln b, borrowing the product law and applying it to a sum.

Don't do this:

There is no rule that simplifies ln⁡(a+b)\ln(a + b). The product law only turns a product into a sum: ln⁡(ab)=ln⁡a+ln⁡b\ln(ab) = \ln a + \ln b, never ln⁡(a+b)\ln(a+b).

The correct way:

Leave ln⁡(a+b)\ln(a + b) as it is. Only split a logarithm when the inside is genuinely a product, quotient, or power.

Dropping the +C+C or the absolute value in ∫1x,dx\int \frac{1}{x},dx.

Where it slips in:

A student writes ∫1x,dx=ln⁡x\int \frac{1}{x},dx = \ln x, missing both the constant of integration and the absolute-value bars.

Don't do this:

Do not write ln⁡x\ln x alone. Without +C+C the antiderivative is incomplete, and without the bars it is undefined for negative xx.

The correct way:

Always write ∫1x,dx=ln⁡∣x∣+C\int \frac{1}{x},dx = \ln|x| + C. The bars keep it valid on both sides of zero, and the +C+C belongs on every indefinite integral.

Forgetting the inner derivative in the chain rule.

Where it slips in:

Differentiating ln⁡(3x2+1)\ln(3x^{2}+1), a student writes 13x2+1\frac{1}{3x^{2}+1} and stops, leaving out the derivative of the inside.

Don't do this:

Do not treat ddx[ln⁡(g(x))]\frac{d}{dx}[\ln(g(x))] as just 1g(x)\frac{1}{g(x)}. That is only half of the chain rule.

The correct way:

Multiply by the inner derivative: ddx[ln⁡(g(x))]=g′(x)g(x)\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}. For ln⁡(3x2+1)\ln(3x^{2}+1) the answer is 6x3x2+1\frac{6x}{3x^{2}+1}.

Confusing ln⁡(x2)\ln(x^{2}) with (ln⁡x)2(\ln x)^{2}.

Where it slips in:

A student treats "ln⁡x\ln x squared" as if the two expressions were interchangeable when applying the power law.

Don't do this:

Do not write (ln⁡x)2=2ln⁡x(\ln x)^{2} = 2\ln x. The power law applies to the exponent inside the logarithm, not to the whole logarithm raised to a power.

The correct way:

Keep them separate: ln⁡(x2)=2ln⁡x\ln(x^{2}) = 2\ln x (exponent inside comes out front), while (ln⁡x)2(\ln x)^{2} is the logarithm multiplied by itself and does not simplify.

Practice Problems On The Natural Logarithm

Answers follow each problem; every derivative and integral checks against the rules above.

  1. Differentiate y=ln⁡(5x)y = \ln(5x).
    (Answer: ln⁡(5x)=ln⁡5+ln⁡x\ln(5x) = \ln 5 + \ln x, so dydx=1x\frac{dy}{dx} = \frac{1}{x}; equivalently 55x=1x\frac{5}{5x} = \frac{1}{x}.)

  2. Differentiate y=ln⁡(x3+2x)y = \ln(x^{3} + 2x).
    (Answer: dydx=3x2+2x3+2x\frac{dy}{dx} = \frac{3x^{2} + 2}{x^{3} + 2x}.)

  3. Differentiate y=xln⁡xy = x\ln x.
    (Answer: product rule gives dydx=ln⁡x+x⋅1x=ln⁡x+1\frac{dy}{dx} = \ln x + x\cdot\frac{1}{x} = \ln x + 1.)

  4. Evaluate ∫3x,dx\displaystyle\int \frac{3}{x},dx.
    (Answer: 3ln⁡∣x∣+C3\ln|x| + C.)

  5. Evaluate ∫1e21x,dx\displaystyle\int_{1}^{e^{2}} \frac{1}{x},dx.
    (Answer: ln⁡(e2)−ln⁡1=2−0=2\ln(e^{2}) - \ln 1 = 2 - 0 = 2.)

  6. Use change of base to find log⁡210\log_{2} 10 to four decimal places.
    (Answer: ln⁡10ln⁡2=2.3025850.693147=3.3219\frac{\ln 10}{\ln 2} = \frac{2.302585}{0.693147} = 3.3219.)

Where Should You Go Next After The Natural Logarithm?

The natural logarithm sits at the junction of algebra and calculus, and several doors open from here.

  1. Logarithmic functions. The algebra side: graphs, domains, and how ln⁡\ln fits the wider family of logarithmic functions.

  2. Logarithm rules. Drill the product, quotient, and power laws that this page proved from the integral.

  3. Calculus. Step back to the big picture of derivatives and integrals, where ln⁡\ln and 1x\frac{1}{x} first meet.

If your child is building these foundations, a live Bhanzu trainer teaches the natural logarithm from the "why" (the area under 1t\tfrac{1}{t} and the growth it measures) in the Bhanzu math classes.

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Frequently Asked Questions

What is the natural logarithm in simple terms?
The natural logarithm of a positive number xx is the power that e≈2.71828e \approx 2.71828 must be raised to in order to give xx, written ln⁡x=log⁡ex\ln x = \log_e x. In calculus it is also the area under the curve y=1ty = \tfrac{1}{t} from t=1t = 1 to t=xt = x.
What is the derivative of the natural logarithm?
It is ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x} for x>0x > 0, and ddx[ln⁡∣x∣]=1x\frac{d}{dx}[\ln|x|] = \frac{1}{x} for all x≠0x \neq 0. When the logarithm wraps another function, the chain rule gives ddx[ln⁡(g(x))]=g′(x)g(x)\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}.
What is the difference between ln and log?
The natural logarithm ln⁡\ln has base ee, while "log⁡\log" usually means base 1010 (the common logarithm). They are related by change of base: log⁡10x=ln⁡xln⁡10\log_{10} x = \frac{\ln x}{\ln 10}.
Why is the integral of 1/x equal to ln|x| + C?
Because the power rule ∫xn,dx=xn+1n+1+C\int x^{n},dx = \frac{x^{n+1}}{n+1}+C fails at n=−1n = -1 (it would divide by zero), and ln⁡∣x∣\ln|x| is the function whose derivative is exactly 1x\frac{1}{x}. The absolute value extends it to negative xx, and every indefinite integral carries a +C+C.
Can you take the natural logarithm of a negative number or zero?
Not in the real numbers. The natural logarithm is defined only for x>0x > 0, because eye^{y} is always positive, so no real power of ee gives zero or a negative result. As x→0+x \to 0^{+}, ln⁡x→−∞\ln x \to -\infty.
Why is it called the "natural" logarithm?
Because base ee is the one base that makes calculus clean: exe^{x} is its own derivative, and ln⁡x\ln x has the simple derivative 1x\frac{1}{x} with no extra constant. Any other base introduces a stray factor of ln⁡b\ln b.
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