What Is The Logarithmic Derivative?
The logarithmic derivative of a differentiable function $f$, defined wherever $f(x) \neq 0$, is the derivative of its natural logarithm:
$$\frac{d}{dx}\big[\ln f(x)\big] = \frac{f'(x)}{f(x)}$$
This follows directly from the chain rule applied to $\ln f(x)$: the outside derivative of $\ln u$ is $\frac{1}{u}$, and the inside derivative is $f'(x)$, so the two multiply to give $\frac{f'(x)}{f(x)}$. When $f$ can be negative, the same formula holds using $\ln|f(x)|$, since $\frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}$ as well.
The quantity $\frac{f'(x)}{f(x)}$ is the relative rate of change: not how fast $f$ is changing in absolute terms, but how fast it is changing as a fraction of its current size. A derivative $f'(x)$ of $200$ means little until you know whether $f(x)$ is $10$ or $10{,}000$. Dividing by $f(x)$ turns the raw rate into a percentage-per-unit rate, which is why the logarithmic derivative is the natural language of growth and decay.
What Is Logarithmic Differentiation?
Logarithmic differentiation is the method that uses the logarithmic derivative to find $f'(x)$ for functions that are painful to differentiate directly. The idea is short: instead of differentiating $y$, differentiate $\ln y$, because the logarithm flattens products into sums using the laws of logarithms.
The method has five steps, and the same five work every time:
Take the natural log of both sides. Start from $y = f(x)$ and write $\ln y = \ln f(x)$.
Expand with log laws. Turn every product into a sum, every quotient into a difference, and every power into a coefficient.
Differentiate both sides implicitly. The left side becomes $\frac{1}{y}\cdot y'$, because $y$ is a function of $x$.
Solve for $y'$. Multiply both sides by $y$.
Resubstitute. Replace $y$ with the original $f(x)$ so the answer is in terms of $x$ only.
Step 2 is where the whole payoff lives. The three log laws below are the engine of the technique, and they are worth knowing cold.
Table: How the three logarithm laws turn a tangled function into an easy sum before you differentiate.
Logarithm law | What it does | Example |
|---|---|---|
$\ln(ab) = \ln a + \ln b$ | product becomes a sum | $\ln\big(x^{2}\sqrt{x-1}\big) = 2\ln x + \tfrac{1}{2}\ln(x-1)$ |
$\ln\frac{a}{b} = \ln a - \ln b$ | quotient becomes a difference | $\ln\frac{x^{3}}{2x+3} = 3\ln x - \ln(2x+3)$ |
$\ln(a^{n}) = n\ln a$ | power becomes a coefficient | $\ln\big((x^2+1)^{3}\big) = 3\ln(x^2+1)$ |
The move that makes logarithmic differentiation feel like magic is implicit differentiation in step 3. Because $y$ depends on $x$, the derivative of $\ln y$ is not $\frac{1}{y}$, it is $\frac{1}{y}\cdot y'$. Forgetting that extra $y'$ factor is the single most common slip, and the mistakes section returns to it.
How Do You Differentiate $x^{x}$ Using Logarithmic Differentiation?
The function $y = x^{x}$ (for $x > 0$) is the headline case, because it has the variable in both the base and the exponent. The power rule needs a constant exponent; the exponential rule needs a constant base. Neither applies, so logs are the way in.
Example 1: Differentiate $y = x^{x}$.
Take the natural log of both sides:
$$\ln y = \ln\big(x^{x}\big) = x\ln x$$
Differentiate both sides with respect to $x$. The left side uses the chain rule; the right side uses the product rule:
$$\frac{1}{y}\cdot y' = (1)\ln x + x\cdot\frac{1}{x} = \ln x + 1$$
Solve for $y'$ by multiplying through by $y$, then resubstitute $y = x^{x}$:
$$y' = y(\ln x + 1) = x^{x}(\ln x + 1)$$
Final answer: $y' = x^{x}(\ln x + 1)$.
An independent check confirms it. Rewrite $x^{x} = e^{x\ln x}$, then differentiate with the chain rule: $\frac{d}{dx}e^{x\ln x} = e^{x\ln x}\cdot(\ln x + 1) = x^{x}(\ln x + 1)$. The two methods agree.
How Do You Differentiate A Big Product Or Quotient?
The second job logarithmic differentiation does well is a stack of products, quotients, and powers, where the product rule and quotient rule would nest into a long, error-prone mess.
Example 2: Differentiate $y = \dfrac{(x^{2}+1)^{3}\sqrt{x-1}}{(2x+3)^{4}}$.
Take logs and expand every factor into a separate term:
$$\ln y = 3\ln(x^{2}+1) + \tfrac{1}{2}\ln(x-1) - 4\ln(2x+3)$$
Differentiate term by term. Each piece is a simple logarithmic derivative:
$$\frac{1}{y}\cdot y' = 3\cdot\frac{2x}{x^{2}+1} + \frac{1}{2}\cdot\frac{1}{x-1} - 4\cdot\frac{2}{2x+3}$$
$$\frac{1}{y}\cdot y' = \frac{6x}{x^{2}+1} + \frac{1}{2(x-1)} - \frac{8}{2x+3}$$
Multiply by $y$ and resubstitute the original expression:
$$y' = \frac{(x^{2}+1)^{3}\sqrt{x-1}}{(2x+3)^{4}}\left(\frac{6x}{x^{2}+1} + \frac{1}{2(x-1)} - \frac{8}{2x+3}\right)$$
Final answer: the boxed product above. Notice there is no nested product-and-quotient rule anywhere; every hard term collapsed into a single fraction the moment the log split it off.
How Do You Differentiate $(\sin x)^{x}$?
Any function of the form (function) raised to (function) yields to the same treatment. Here the base is $\sin x$ and the exponent is $x$, so the method is identical to $x^{x}$.
Example 3: Differentiate $y = (\sin x)^{x}$, on an interval where $\sin x > 0$.
$$\ln y = x\ln(\sin x)$$
Differentiate the right side with the product rule, and use $\frac{d}{dx}\ln(\sin x) = \frac{\cos x}{\sin x} = \cot x$:
$$\frac{1}{y}\cdot y' = (1)\ln(\sin x) + x\cdot\cot x$$
$$y' = (\sin x)^{x}\big(\ln(\sin x) + x\cot x\big)$$
Final answer: $y' = (\sin x)^{x}\big(\ln(\sin x) + x\cot x\big)$.
The $\cot x$ term is exactly the logarithmic derivative of $\sin x$, since $\frac{(\sin x)'}{\sin x} = \frac{\cos x}{\sin x}$. Every log-based derivative you meet is built from these small $\frac{f'}{f}$ blocks, which is why knowing the differentiation of trigonometric functions pays off here.
Why Does Logarithmic Differentiation Work?
The method is trustworthy because it does not invent new rules. It only reroutes the product, quotient, and power rules through a logarithm. The clearest way to see this is to watch it rebuild a rule you already trust.
Take two positive functions $f$ and $g$, and let $y = f(x)g(x)$. Then:
Taking logs gives $\ln y = \ln f + \ln g$, a sum instead of a product.
Differentiating gives $\frac{y'}{y} = \frac{f'}{f} + \frac{g'}{g}$, a sum of two logarithmic derivatives.
Multiplying by $y = fg$ gives $y' = fg\left(\frac{f'}{f} + \frac{g'}{g}\right) = f'g + fg'$.
That last line is the ordinary product rule, recovered exactly. The same routine rebuilds the power rule: for $y = x^{n}$, $\ln y = n\ln x$, so $\frac{y'}{y} = \frac{n}{x}$, giving $y' = x^{n}\cdot\frac{n}{x} = nx^{n-1}$.
Geometrically, the logarithmic derivative is the slope of the graph of $\ln f$ against $x$. On ordinary axes a doubling from $10$ to $20$ and a doubling from $1000$ to $2000$ look wildly different; on a log scale they have the same proportional rise, so they share the same slope. That is the visual meaning of $\frac{f'}{f}$: constant relative growth is a straight line on log paper, and its steepness is the logarithmic derivative.
How Does The Logarithmic Derivative Connect To Integration?
Reading the definition backwards turns it into one of the most useful integral patterns in calculus. If differentiating $\ln|f(x)|$ gives $\frac{f'(x)}{f(x)}$, then integrating $\frac{f'(x)}{f(x)}$ must return $\ln|f(x)|$:
$$\int \frac{f'(x)}{f(x)},dx = \ln|f(x)| + C$$
Whenever the top of a fraction is the derivative of the bottom, the integral is a natural log. For instance, $\int \frac{2x}{x^{2}+1},dx = \ln(x^{2}+1) + C$, because $2x$ is the derivative of $x^{2}+1$. The logarithmic derivative and this integral are the same statement read in two directions, which is why spotting an $\frac{f'}{f}$ shape is a shortcut both when differentiating and when integrating.
Who Invented The Logarithmic Derivative?
Logarithms arrived first as a labour-saving device, decades before calculus existed to give them a derivative. Their inventor spent twenty years building the tables by hand so that other people would not have to.
Two later figures turned the logarithm into a derivative:
Gottfried Wilhelm Leibniz (1646–1716, Germany) built the differential calculus and the $d$-notation that lets us write $\frac{d}{dx}\ln f$ at all, and studied the natural logarithm as the area under $\frac{1}{x}$.
Johann Bernoulli (1667–1748, Switzerland), one of Leibniz's closest correspondents, worked out how to differentiate expressions with a variable exponent, such as $x^{x}$, which is precisely the case logarithmic differentiation was made for.
Where Is The Logarithmic Derivative Used In The Real World?
The relative rate of change is the quantity that many fields actually care about, so the logarithmic derivative appears far beyond a calculus exam.
Economics and finance: continuously compounded growth, inflation, and returns are all relative rates, so $\frac{f'}{f}$ is the instantaneous growth rate of a price, an investment, or a GDP figure.
Biology and medicine: populations of bacteria, tumours, and species grow in proportion to their current size, which is a differential equation written directly in terms of the logarithmic derivative.
Physics and engineering: radioactive decay and RC-circuit discharge obey a constant negative logarithmic derivative, giving the exponential decay curve and its half-life.
Chemistry: reaction rate laws and pH, which is itself a base-10 logarithm of concentration, lean on the same relative-rate idea.
Data science: log-likelihood optimisation differentiates $\ln L$ rather than $L$, using the logarithmic derivative to find the parameters that best fit data.
Across every one of these, the shared question is the same, how fast is this changing compared to how big it already is, and that ratio is the logarithmic derivative.
What Are The Most Common Mistakes With The Logarithmic Derivative?
These three errors account for most lost marks on logarithmic differentiation problems, and they match the documented student mistakes on the derivative of $x^{x}$ and the recurring "why multiply by $y$" questions readers ask.
Treating $y$ as a constant when differentiating $\ln y$.
Where it slips in:
A student writes $\frac{d}{dx}\ln y = \frac{1}{y}$ and stops, forgetting that $y$ is a function of $x$.
Don't do this:
Do not drop the chain-rule factor. $\frac{1}{y}$ alone is the derivative only if $y$ is a plain variable, not a function.
The correct way:
Differentiate implicitly: $\frac{d}{dx}\ln y = \frac{1}{y}\cdot y'$. That extra $y'$ is the whole reason the method can solve for the derivative.
Splitting the logarithm of a sum.
Where it slips in:
Facing $\ln(x^{2}+1)$, a student writes it as $\ln(x^{2}) + \ln(1)$ or $2\ln x + \ln 1$, inventing a law that does not exist.
Don't do this:
Do not break $\ln(a+b)$ into $\ln a + \ln b$. The product law applies to $\ln(ab)$, never to a sum inside the logarithm.
The correct way:
Leave $\ln(x^{2}+1)$ intact and differentiate it as a single logarithmic derivative: $\frac{d}{dx}\ln(x^{2}+1) = \frac{2x}{x^{2}+1}$.
Forgetting to multiply back by the original function.
Where it slips in:
A student reaches $\frac{y'}{y} = \ln x + 1$ and reports $\ln x + 1$ as the answer, leaving $y'$ unsolved.
Don't do this:
Do not stop at $\frac{y'}{y}$. That expression is the logarithmic derivative, not the derivative itself.
The correct way:
Multiply both sides by $y$ and resubstitute the original function: $y' = y(\ln x + 1) = x^{x}(\ln x + 1)$.
Practice Problems On The Logarithmic Derivative
Use logarithmic differentiation on each. Answers follow every line, and each has been checked.
Differentiate $y = x^{2x}$.
(Answer: $y' = 2x^{2x}(\ln x + 1)$, from $\ln y = 2x\ln x$.)Differentiate $y = x^{\ln x}$.
(Answer: $y' = x^{\ln x}\cdot\dfrac{2\ln x}{x}$, from $\ln y = (\ln x)^{2}$.)Differentiate $y = x^{1/x}$.
(Answer: $y' = x^{1/x}\cdot\dfrac{1-\ln x}{x^{2}}$.)Differentiate $y = (\cos x)^{x}$ where $\cos x > 0$.
(Answer: $y' = (\cos x)^{x}\big(\ln(\cos x) - x\tan x\big)$.)Differentiate $y = \dfrac{(x+1)^{2}(x-2)^{3}}{x+3}$.
(Answer: $y' = y\left(\dfrac{2}{x+1} + \dfrac{3}{x-2} - \dfrac{1}{x+3}\right)$.)Re-derive the derivative of $y = x^{5}$ with logs.
(Answer: $\ln y = 5\ln x$, so $\dfrac{y'}{y} = \dfrac{5}{x}$ and $y' = 5x^{4}$.)
Where Should You Go Next After The Logarithmic Derivative?
The logarithmic derivative sits at the crossroads of logarithms and calculus, and several doors open from here.
Logarithm rules. The product, quotient, and power laws that power step 2 of the method, with more worked expansions.
Differentiation of trigonometric functions. The small $\frac{f'}{f}$ blocks (like $\cot x$ from $\sin x$) that show up inside every trig-based logarithmic derivative.
Integration. The reverse direction, where the $\frac{f'}{f}$ pattern becomes $\ln|f| + C$.
If your child is meeting these ideas in India's NCERT Class 12 continuity-and-differentiability chapter or in UK A-Level Mathematics, a live Bhanzu trainer teaches logarithmic differentiation starting from the "why" behind $\frac{f'}{f}$ in the Bhanzu math tutoring program.
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