Derivatives of ln x and logₐ(x): Worked examples

The scenario matrix

Before working anything, let us list every case-class this topic can generate. Each row is a "cell." The worked examples below are tagged with the cell they cover, so together they fill the whole grid.

#	Cell (case class)	What makes it different	Example
C1	Plain $ln$ of a polynomial	pure chain rule $u^{'} / u$	(a)
C2	Non- $e$ base $lo g_{a}$	extra $\frac{1}{ln a}$ factor	(b)
C3	Simplify-first (log laws)	avoid chain rule entirely	(c)
C4	Negative / absolute-value input	domain, $ln ∣ x ∣$	(d)
C5	Product/quotient inside	log laws turn it into a sum	(e)
C6	Variable in the exponent	logarithmic differentiation	(f)
C7	Real-world rate (word problem)	units, interpretation	(g)
C8	Exam twist / degenerate	limiting value, hidden constant	(h)

We will hit all eight cells with eight examples.

Reading the picture first

The single fact behind every example is the shape of $ln x$ and its slope $1/ x$ . Look at the figure: the height climbs but ever more slowly, and the little slope triangles get flatter as you move right.

Figure — Derivatives of ln x and logₐ(x)

The worked examples

(a) Cell C1 — plain $ln$ of a polynomial

(b) Cell C2 — non- $e$ base

(c) Cell C3 — simplify first with log laws

(d) Cell C4 — negative input and $ln ∣ x ∣$

Worked example Differentiate

$p (x) = ln ∣ x ∣$ , and evaluate $p^{'} (- 2)$ Forecast: $ln$ of a negative number is undefined — but $ln ∣ x ∣$ is fine for $x < 0$ . Do you think the slope formula flips sign on the left?

For $x > 0$ : $∣ x ∣ = x$ , so $p (x) = ln x$ and $p^{'} (x) = \frac{1}{x}$ . Why this step? No absolute value to worry about on the positive side.
For $x < 0$ : $∣ x ∣ = - x$ , so $p (x) = ln (- x)$ . Chain rule with $u = - x$ , $u^{'} = - 1$ : $p^{'} (x) = \frac{u ^{'}}{u} = \frac{- 1}{- x} = \frac{1}{x} .$ Why this step? The two minus signs cancel — the slope is $1/ x$ on both sides.
Combine: $p^{'} (x) = \frac{1}{x}$ for all $x \neq = 0$ . Why this step? One tidy rule covers the whole domain minus the forbidden point $x = 0$ .

Verify: $p^{'} (- 2) = \frac{1}{- 2} = - 0.5$ . The picture in the figure below shows the mirror-image branch; on the left the curve falls as $x$ increases toward $0$ , so a negative slope is right. ✓

(e) Cell C5 — product inside the log

(f) Cell C6 — variable in the exponent (logarithmic differentiation)

Worked example Differentiate

$y = (2 x)^{x}$ for $x > 0$ Forecast: the exponent is a variable, so neither the power rule nor $a^{x}$ rule applies alone. What tool pulls an exponent down? Logs. See Logarithmic Differentiation.

Take $ln$ of both sides: $ln y = x ln (2 x)$ . Why this step? $ln A^{p} = p ln A$ drags the variable exponent $x$ into a product we can differentiate.
Simplify the right: $ln (2 x) = ln 2 + ln x$ , so $ln y = x ln 2 + x ln x$ . Why this step? Splitting first makes the product rule cleaner.
Differentiate both sides (Implicit Differentiation on the left): $\frac{y ^{'}}{y} = ln 2 + (ln x + x \cdot \frac{1}{x}) = ln 2 + ln x + 1.$ Why this step? Left side: $\frac{d}{d x} ln y = \frac{y ^{'}}{y}$ (chain rule). Right: product rule on $x ln x$ .
Solve for $y^{'}$ : $y^{'} = (2 x)^{x} (ln 2 + ln x + 1) .$ Why this step? Multiply back by $y = (2 x)^{x}$ .

Verify: note $ln 2 + ln x = ln (2 x)$ , so $y^{'} = (2 x)^{x} (ln (2 x) + 1)$ — the same shape as the parent's $x^{x} (ln x + 1)$ , with $x \to 2 x$ inside the log. At $x = 1$ : $y = 2^{1} = 2$ and $y^{'} = 2 (ln 2 + 0 + 1) = 2 (ln 2 + 1) \approx 3.386$ . ✓

(g) Cell C7 — real-world rate (word problem)

Worked example A quiet zone's loudness in decibels is

$L (I) = 10 lo g_{10} (\frac{I}{I _{0}})$ , where $I$ is sound intensity and $I_{0}$ is a fixed reference. How fast does $L$ change with $I$ ? Evaluate at $I = 100 I_{0}$ . Forecast: more intensity means more loudness — but the log means each extra unit of $I$ adds less. Guess: will $\frac{d L}{d I}$ grow or shrink as $I$ grows?

Rewrite: $L = 10 (lo g_{10} I - lo g_{10} I_{0})$ ; the second term is constant. Why this step? $lo g_{10} I_{0}$ has zero derivative — only $lo g_{10} I$ varies.
Differentiate using the master rule $\frac{d}{d I} lo g_{10} I = \frac{1}{I l n 10}$ : $\frac{d L}{d I} = 10 \cdot \frac{1}{I l n 10} = \frac{10}{I l n 10} .$ Why this step? Base-10 log carries the $\frac{1}{l n 10}$ factor (Cell C2 lesson).
Evaluate at $I = 100 I_{0}$ : $\frac{d L}{d I} = \frac{10}{100 I _{0} ln 10} = \frac{1}{10 I _{0} ln 10} .$ Why this step? Plug the value in and keep $I_{0}$ symbolic (it is a fixed constant with units of intensity).

Verify: units are $decibels per unit intensity$ , and the answer shrinks like $1/ I$ — exactly the flattening from figure s01. Louder sounds need a bigger jump in intensity to gain the same decibel, which is why decibels are logarithmic. Numerically $\frac{10}{100 l n 10} \approx 0.04343$ dB per unit (taking $I_{0} = 1$ ). ✓

(h) Cell C8 — exam twist / degenerate case

Worked example Find

$\frac{d}{d x} lo g_{a} x$ and state what happens as the base $a \to 1^{+}$ . Forecast: the formula is $\frac{1}{x l n a}$ . What does $ln a$ do as $a$ slides down toward $1$ ? That is the trap.

Master formula: $\frac{d}{d x} lo g_{a} x = \frac{1}{x ln a}$ . Why this step? Direct from change of base; $ln a$ is a constant for fixed $a$ .
Degenerate limit: as $a \to 1^{+}$ , $ln a \to 0^{+}$ , so $\frac{1}{x ln a} \to + \infty$ . Why this step? Dividing by something shrinking to $0$ blows the slope up.
Interpret: $lo g_{1} x$ is undefined — that is why the parent note demands $a \neq = 1$ . The derivative diverging signals the forbidden base. Why this step? Calculus warns you at the boundary: an infinite slope means the function itself has collapsed.

Verify: sanity check the other special base $a = e$ : $ln e = 1$ gives $\frac{1}{x \cdot 1} = \frac{1}{x}$ , recovering plain $ln$ (parent §3). And $a = 2$ at $x = 1$ gives $\frac{1}{1 \cdot l n 2} \approx 1.4427$ . Both finite and correct; only $a = 1$ misbehaves. ✓

Active recall

Recall Which cell does each problem hit? (cover the right side)

$\frac{d}{d x} ln (x^{2} + 4)$ ::: C1 — plain chain rule, answer $\frac{2 x}{x ^{2} + 4}$ . $\frac{d}{d x} lo g_{10} (7 x)$ ::: C2 — extra $\frac{1}{l n 10}$ , answer $\frac{1}{x l n 10}$ . $\frac{d}{d x} ln \frac{x ^{3}}{x + 1}$ ::: C3 — split first, answer $\frac{3}{x} - \frac{1}{2 ( x + 1 )}$ . $\frac{d}{d x} ln ∣ x ∣$ ::: C4 — $\frac{1}{x}$ for all $x \neq = 0$ . $\frac{d}{d x} ln (x^{2} e^{3 x})$ ::: C5 — $\frac{2}{x} + 3$ . $\frac{d}{d x} (2 x)^{x}$ ::: C6 — $(2 x)^{x} (ln (2 x) + 1)$ . Why does $lo g_{a} x$ break as $a \to 1$ ? ::: C8 — $ln a \to 0$ , slope $\to \infty$ , base $1$ forbidden.

Connections

Chain Rule — the engine behind C1, C4.
Logarithm Laws — turns C3, C5 into sums.
Logarithmic Differentiation — the only route for C6.
Implicit Differentiation — used inside C6 on $\frac{d}{d x} ln y$ .
Derivative of e^x and a^x — the inverse fact simplifying $ln e^{3 x}$ in C5.
Standard Limit (1+t)^{1/t} → e — the foundation the parent's $1/ x$ rests on.
Parent topic — the master formulas this page exercises.