What Are The Derivatives Of Logarithmic Functions?
The derivatives of logarithmic functions are the rules for differentiating $\ln x$, $\log_a x$, and any logarithm of a function. There are three you need, and the last two are built from the first.
$$\frac{d}{dx}\ln x = \frac{1}{x}, \qquad \frac{d}{dx}\log_a x = \frac{1}{x\ln a}, \qquad \frac{d}{dx}\ln\big(u(x)\big) = \frac{u'(x)}{u(x)}$$
Read them in plain words. The natural logarithm $\ln x$ (base $e$) has the cleanest derivative in all of calculus: $\frac{1}{x}$. A logarithm to any other base $a$ carries one extra constant factor, $\frac{1}{\ln a}$. And when the logarithm wraps a whole function $u(x)$, the chain rule divides the derivative of the inside by the inside itself.
Two conditions keep the formulas honest. The argument of a real logarithm must be positive, so $\frac{d}{dx}\ln x = \frac{1}{x}$ holds for $x > 0$; the base $a$ must satisfy $a > 0$ and $a \neq 1$, so that $\ln a \neq 0$ and the division is defined. For a refresher on the function itself before differentiating it, see the natural logarithm page and the algebra recap of logarithm rules.
How Do You Prove The Derivative Of The Natural Logarithm?
The shortest proof runs backward through the exponential. It uses the fact that $\ln x$ is the inverse of $e^x$, so it pairs naturally with implicit differentiation and the derivative of the exponential function.
Let $y = \ln x$. By the definition of the natural logarithm, this is the same statement as
$$e^{y} = x.$$
Differentiate both sides with respect to $x$. The left side needs the chain rule because $y$ is a function of $x$:
$$e^{y},\frac{dy}{dx} = 1.$$
Solve for $\frac{dy}{dx}$ and replace $e^{y}$ with $x$:
$$\frac{dy}{dx} = \frac{1}{e^{y}} = \frac{1}{x}.$$
That is the whole proof. Because $e^{y} = x$ is only true for $x > 0$, the derivative $\frac{1}{x}$ inherits the same domain.
The first-principle route reaches the same place from the limit definition of the derivative. For $f(x) = \ln x$,
$$f'(x) = \lim_{h \to 0}\frac{\ln(x+h) - \ln x}{h} = \lim_{h \to 0}\frac{1}{h}\ln\left(1 + \frac{h}{x}\right) = \frac{1}{x}\lim_{h \to 0}\ln\left(1 + \frac{h}{x}\right)^{x/h}.$$
As $h \to 0$, the inside quantity $\left(1 + \frac{h}{x}\right)^{x/h}$ approaches $e$, so the limit is $\frac{1}{x}\ln e = \frac{1}{x}$. Both proofs agree, and both explain why $e$ is the base that makes the derivative simplest.
What Is The Derivative Of Log Base A?
A logarithm to base $a$ is just the natural logarithm rescaled, so its derivative needs no new proof. Use the change-of-base identity:
$$\log_a x = \frac{\ln x}{\ln a}.$$
Here $\ln a$ is a constant, so differentiating pulls it straight out:
$$\frac{d}{dx}\log_a x = \frac{1}{\ln a}\cdot\frac{d}{dx}\ln x = \frac{1}{\ln a}\cdot\frac{1}{x} = \frac{1}{x\ln a}.$$
Two special cases are worth fixing in memory. When $a = e$, $\ln a = \ln e = 1$, and the formula collapses back to $\frac{1}{x}$. When $a = 10$, the common logarithm written plainly as $\log x$ has derivative
$$\frac{d}{dx}\log_{10} x = \frac{1}{x\ln 10} \approx \frac{1}{2.3026,x}.$$
That extra $\ln 10 \approx 2.3026$ is exactly why $\frac{d}{dx}\ln x$ and $\frac{d}{dx}\log x$ are not the same number, a distinction the mistakes section returns to.
How Does The Chain Rule Extend The Derivatives Of Logarithmic Functions?
Most real problems ask for the logarithm of a function, not a bare $x$. Write $y = \ln\big(u(x)\big)$ and apply the chain rule: the outside derivative is $\frac{1}{u}$, and you multiply by the inside derivative $u'(x)$.
$$\frac{d}{dx}\ln\big(u(x)\big) = \frac{1}{u(x)}\cdot u'(x) = \frac{u'(x)}{u(x)}$$
The pattern is worth memorising as a sentence: the derivative of a log is the derivative of the inside over the inside. For a general base, the same reasoning gives $\frac{d}{dx}\log_a\big(u(x)\big) = \frac{u'(x)}{u(x)\ln a}$.
One domain refinement matters in practice. The function $\ln|x|$ is defined for every $x \neq 0$, and its derivative is still $\frac{1}{x}$:
$$\frac{d}{dx}\ln|x| = \frac{1}{x}, \qquad x \neq 0.$$
This is why integral tables write $\int \frac{1}{x},dx = \ln|x| + C$ with the absolute value, covering negative $x$ as well as positive. A fuller list of these standard results lives in the table of derivatives.
How Do The Derivatives Of Logarithmic Functions Power Logarithmic Differentiation?
The chain-rule form is the engine behind a technique for differentiating awkward products, quotients, and variable exponents. The idea: take the natural logarithm of both sides first, so multiplication becomes addition, then differentiate.
If $y = f(x)$, taking logs and differentiating gives $\frac{y'}{y} = \frac{d}{dx}\ln f(x)$, so
$$y' = y\cdot\frac{d}{dx}\ln f(x) = f(x)\cdot\frac{d}{dx}\ln f(x).$$
That single step turns a nested product into a sum of simple $\frac{u'}{u}$ pieces. It is the only clean way to handle a variable raised to a variable power, such as $x^{x}$, where neither the power rule nor the exponential rule applies on its own. The dedicated method, with more cases, is covered in logarithmic differentiation; the quantity $\frac{f'}{f}$ it produces has its own name and page, the logarithmic derivative.
How Do You Differentiate Logarithmic Functions? Worked Examples
Each example is fully stepped, and each answer is checked by re-reading the chain-rule pattern $\frac{u'}{u}$ or by differentiating an equivalent expanded form.
Example 1: A logarithm of a polynomial (chain rule).
Differentiate $y = \ln(x^{2} + 1)$.
Here $u = x^{2} + 1$, so $u' = 2x$. Apply $\frac{u'}{u}$:
$$\frac{dy}{dx} = \frac{2x}{x^{2} + 1}$$
Check: the inside $x^{2} + 1$ is always positive, so the log is defined everywhere, and the denominator never vanishes. The sign of the derivative matches the graph, negative for $x < 0$, zero at $x = 0$, positive for $x > 0$.
Final answer: $\dfrac{dy}{dx} = \dfrac{2x}{x^{2} + 1}$.
Example 2: A general base with the chain rule.
Differentiate $y = \log_{2}(3x + 1)$.
Combine the base formula with the chain rule: divide $u' = 3$ by $u\ln a = (3x + 1)\ln 2$.
$$\frac{dy}{dx} = \frac{3}{(3x + 1)\ln 2}$$
Check at $x = 1$: the slope is $\frac{3}{4\ln 2} \approx \frac{3}{2.7726} \approx 1.0821$, a positive value, consistent with the increasing curve.
Final answer: $\dfrac{dy}{dx} = \dfrac{3}{(3x + 1)\ln 2}$.
Example 3: A variable exponent (logarithmic differentiation).
Differentiate $y = x^{x}$ for $x > 0$.
Take the natural logarithm of both sides, then use $\ln(x^{x}) = x\ln x$:
$$\ln y = x\ln x$$
Differentiate both sides. The right side needs the product rule ($\frac{d}{dx}[x\ln x] = \ln x + x\cdot\frac{1}{x}$):
$$\frac{y'}{y} = \ln x + 1$$
Multiply through by $y = x^{x}$:
$$y' = x^{x}\big(\ln x + 1\big)$$
Check at $x = 1$: $y' = 1^{1}(\ln 1 + 1) = 1\cdot(0 + 1) = 1$, which matches the known slope of $x^{x}$ at $x = 1$.
Final answer: $y' = x^{x}\big(\ln x + 1\big)$.
Example 4: Expand before differentiating.
Differentiate $y = \ln\left(\dfrac{x^{2}\sin x}{2x + 1}\right)$.
Rather than fight one large chain rule, use log rules to split the expression first:
$$y = 2\ln x + \ln(\sin x) - \ln(2x + 1)$$
Now differentiate term by term, each a simple $\frac{u'}{u}$:
$$\frac{dy}{dx} = \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{2}{2x + 1} = \frac{2}{x} + \cot x - \frac{2}{2x + 1}$$
The middle term uses the derivatives of trigonometric functions, where $\frac{d}{dx}\ln(\sin x) = \frac{\cos x}{\sin x}$.
Final answer: $\dfrac{dy}{dx} = \dfrac{2}{x} + \cot x - \dfrac{2}{2x + 1}$.
What Are The Standard Derivatives Of Logarithmic Functions?
These are the results worth knowing on sight. Every one follows from $\frac{d}{dx}\ln x = \frac{1}{x}$ plus the chain rule or change of base.
Table: Standard derivatives of logarithmic functions and their conditions.
Function | Derivative | Condition |
|---|---|---|
$\ln x$ | $\dfrac{1}{x}$ | $x > 0$ |
$\ln | x | $ |
$\log_a x$ | $\dfrac{1}{x\ln a}$ | $x > 0,; a > 0,; a \neq 1$ |
$\log_{10} x$ | $\dfrac{1}{x\ln 10}$ | $x > 0$ |
$\ln\big(u(x)\big)$ | $\dfrac{u'(x)}{u(x)}$ | $u(x) > 0$ |
$\log_a\big(u(x)\big)$ | $\dfrac{u'(x)}{u(x)\ln a}$ | $u(x) > 0,; a \neq 1$ |
For where these sit among the power, product, and quotient rules, see the wider rules of differentiation.
Why Do The Derivatives Of Logarithmic Functions Work?
The formula $\frac{1}{x}$ can feel like a coincidence until you read it as a slope and as a rate.
The slope shrinks because the curve slows. The graph of $y = \ln x$ rises steeply just past $x = 0$ and flattens as $x$ grows. A tangent drawn at $x = 1$ has slope $1$; at $x = 4$ it has slope $\frac{1}{4}$; at $x = 100$ it is nearly flat. The derivative $\frac{1}{x}$ is precisely that shrinking positive slope, which is why the logarithm keeps rising forever yet never levels off.
It is the inverse of exponential growth. Because $\ln x$ undoes $e^{x}$, its rate of change is the reciprocal of the exponential's rate. Where $e^{x}$ climbs faster and faster, its mirror image climbs slower and slower, and the reciprocal $\frac{1}{x}$ captures that reflection exactly.
The chain-rule form is a relative rate. The quantity $\frac{u'}{u}$ is a change measured against the current size. Growing by $3$ units matters more when you have $10$ than when you have $1000$, and $\frac{u'}{u}$ is what expresses that proportional view, the reason logarithms model percentage growth so well.
Seen together, the algebra ($\frac{1}{x}$) and the geometry (a positive, shrinking tangent slope) are two readings of one fact: the logarithm accumulates change at a rate inversely proportional to where it already is.
Who Discovered The Derivatives Of Logarithmic Functions?
Logarithms were a calculation tool for decades before calculus existed to differentiate them, and the link $\frac{d}{dx}\ln x = \frac{1}{x}$ was first spotted as a fact about areas, not slopes.
Two more figures shaped the story:
John Napier (1550–1617, Scotland) invented logarithms in 1614 as a device for turning multiplication into addition, decades before anyone asked about their rate of change.
Leonhard Euler (1707–1783, Switzerland) placed the constant $e$ at the centre of the subject and made the natural logarithm the default, which is what makes $\frac{d}{dx}\ln x = \frac{1}{x}$ the base case every other logarithm is measured against.
Where Are The Derivatives Of Logarithmic Functions Used In The Real World?
Because the chain-rule form $\frac{u'}{u}$ is a proportional rate of change, logarithmic derivatives appear wherever quantities are measured on a ratio or percentage scale.
Chemistry: pH is defined as $-\log_{10}[\text{H}^{+}]$, so the rate a solution acidifies as ion concentration changes is a logarithmic derivative.
Acoustics and seismology: decibels and the Richter scale are logarithmic, and how loudness or magnitude responds to a change in energy is exactly a $\frac{1}{x\ln a}$ relationship.
Economics and finance: the derivative of $\ln(\text{price})$ is the continuously compounded return, and $\frac{f'}{f}$ measures elasticity, how a percentage change in one quantity drives a percentage change in another.
Information theory: entropy is built from $\log$ terms, and differentiating them is how compression and coding limits are optimised.
Biology and physics: solving growth and decay models, from populations to radioactive half-lives, needs the natural logarithm, and its $\frac{1}{x}$ derivative is what lets those equations be inverted.
One derivative rule, read as a proportional rate, ties together acidity, loudness, money, and information, fields that share nothing except a logarithmic scale.
What Are The Most Common Mistakes With Derivatives Of Logarithmic Functions?
These four errors account for most lost marks on this topic, and each matches a question real students ask on r/calculus, r/learnmath, and university error handouts.
Dropping the inner derivative in the chain rule.
Where it slips in:
A student writes $\dfrac{d}{dx}\ln(x^{2} + 1) = \dfrac{1}{x^{2} + 1}$ and stops, treating the inside as if it were a bare $x$.
Don't do this:
Do not forget the $u'(x)$ factor. The derivative of a log is not simply one over the inside.
The correct way:
Multiply by the derivative of the inside: $\dfrac{u'}{u} = \dfrac{2x}{x^{2} + 1}$.
Confusing the derivative of $\ln x$ with the derivative of $\log x$.
Where it slips in:
A student assumes $\dfrac{d}{dx}\log_{10} x = \dfrac{1}{x}$, copying the natural-log result onto a base-$10$ logarithm.
Don't do this:
Do not drop the $\ln a$ factor. Only the natural logarithm has the clean $\frac{1}{x}$ derivative.
The correct way:
Keep the base constant: $\dfrac{d}{dx}\log_{10} x = \dfrac{1}{x\ln 10}$, and in general $\dfrac{1}{x\ln a}$.
Ignoring the domain and the absolute value.
Where it slips in:
A student differentiates $\ln(\text{negative expression})$ without noticing the log is undefined there, or writes $\int \frac{1}{x},dx = \ln x + C$ and misses negative $x$.
Don't do this:
Do not apply $\frac{1}{x}$ blindly. The real logarithm needs a positive argument.
The correct way:
State the domain first. Use $\ln|x|$ when negative inputs are in play, since $\dfrac{d}{dx}\ln|x| = \dfrac{1}{x}$ for $x \neq 0$.
Fighting a big product instead of expanding first.
Where it slips in:
Faced with $\ln\left(\frac{x^{2}\sin x}{2x + 1}\right)$, a student launches a single tangled chain rule and loses track of terms.
Don't do this:
Do not differentiate the whole product at once. It multiplies the chances of a slip.
The correct way:
Use log rules to expand into a sum first: $2\ln x + \ln(\sin x) - \ln(2x + 1)$, then differentiate each simple piece.
Practice Problems On Derivatives Of Logarithmic Functions
Work each one, then check against the answer. Answers are verified.
Differentiate $y = \ln(5x)$.
(Answer: $\ln(5x) = \ln 5 + \ln x$, so $y' = \frac{1}{x}$; equivalently $\frac{5}{5x} = \frac{1}{x}$.)Differentiate $y = \log_{10} x$.
(Answer: $y' = \frac{1}{x\ln 10}$.)Differentiate $y = \ln(x^{3} + 2x)$.
(Answer: $y' = \frac{3x^{2} + 2}{x^{3} + 2x}$.)Differentiate $y = \log_{3}(x^{2})$.
(Answer: $y' = \frac{2x}{x^{2}\ln 3} = \frac{2}{x\ln 3}$.)Differentiate $y = \ln(\cos x)$.
(Answer: $y' = \frac{-\sin x}{\cos x} = -\tan x$.)Differentiate $y = x^{\ln x}$ for $x > 0$.
(Answer: $\ln y = (\ln x)^{2}$, so $\frac{y'}{y} = \frac{2\ln x}{x}$ and $y' = x^{\ln x}\cdot\frac{2\ln x}{x}$.)
Where Should You Go Next After Derivatives Of Logarithmic Functions?
The log derivatives sit inside a small family of results, and several natural doors open from here.
Logarithmic differentiation. The full technique for products, quotients, and variable exponents that the chain-rule form powers.
Derivatives of exponential functions. The mirror topic, since $\ln x$ is the inverse of $e^{x}$ and each proof leans on the other.
The natural logarithm. Go back to the function itself, its base $e$, and the properties the derivative depends on.
If your child is meeting these derivatives for the first time, a live Bhanzu trainer teaches them from the tangent-slope picture up, so the rules feel earned, in the Bhanzu math program.
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