In information theory, Claude Shannon defined entropy as a measure of uncertainty. When analysing how information content changes with probability, we encounter the natural logarithm. The derivative of $\ln x$ being $\frac{1}{x}$ appears in economics (marginal utility), physics (entropy), biology (allometric scaling), and computer science (algorithm analysis). This simple reciprocal rule connects to some of the deepest ideas in science — all because the natural logarithm and the exponential function are inverses.
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If $y = \ln x$, what might $\frac{dy}{dx}$ be? Think about the relationship between $\ln x$ and $e^x$ — they are inverse functions. How does this affect their derivatives?
Core Content
Method 1: Inverse functions
If $y = \ln x$, then $x = e^y$. Differentiate implicitly with respect to $y$:
dx/dy = e^y = x
Therefore:
dy/dx = 1/(dx/dy) = 1/x
Method 2: First principles
f'(x) = lim_{h→0} (ln(x+h) - ln(x))/h = lim_{h→0} (1/h)·ln((x+h)/x) = lim_{h→0} ln((1 + h/x)^{1/h})
Let $t = h/x$, so $h = xt$ and $1/h = 1/(xt)$:
= lim_{t→0} ln((1 + t)^{1/(xt)}) = lim_{t→0} (1/x)·ln((1+t)^{1/t}) = (1/x)·ln(e) = 1/x
since $\lim_{t \to 0} (1+t)^{1/t} = e$.
For composite functions:
d/dx(ln(u)) = (1/u) · du/dx = u'/u
Examples:
Important: $\frac{d}{dx}(\ln(kx)) = \frac{1}{x}$ for any constant $k > 0$. The constant cancels out! This means all curves $y = \ln(kx)$ have the same gradient at each $x$-value.
Find the derivative of $f(x) = x^2 \ln x$.
$f'(x)$ in simplified form.
Use product rule: u = x^2, v = ln(x)
u' = 2x
v' = 1/x
f'(x) = u'v + uv' = 2x·ln(x) + x^2·(1/x)
= 2x ln(x) + x
= x(2 ln(x) + 1)
$f'(x) = x(2\ln x + 1)$.
Find the derivative of $f(x) = \frac{\ln x}{x}$.
Answer:
$f'(x) = \frac{(1/x) \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$.
The natural logarithm $\ln x$ is only defined for $x > 0$. But $\ln|x|$ is defined for all $x \neq 0$:
Therefore:
d/dx(ln|x|) = 1/x for all x ≠ 0
This is useful when integrating (Y12) and when working with functions that can take negative values.
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{d}{dx}(\ln u) = \frac{u'}{u}$
$\frac{d}{dx}(\ln(kx)) = \frac{1}{x}$
$\frac{d}{dx}(\ln|x|) = \frac{1}{x}$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$. This follows from the inverse function relationship: if $y = \ln x$, then $x = e^y$, so $\frac{dx}{dy} = e^y = x$, giving $\frac{dy}{dx} = \frac{1}{x}$. The derivative is a simple reciprocal, but it is only defined for $x > 0$ (the domain of $\ln x$).
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Differentiate $f(x) = x^3 \ln x$. Show all working and simplify. 3 MARKS
9. Find the equation of the tangent to $y = \ln x$ at $x = e$. Give your answer in exact form. 3 MARKS
10. In economics, the utility of wealth $W$ is sometimes modelled as $U = \ln(W)$. A person has $W = 100$. Find the marginal utility $\frac{dU}{dW}$ and interpret its meaning. If a tax reduces their wealth by 10%, calculate the percentage decrease in utility. Explain why this model supports progressive taxation (where wealthier people pay a higher percentage of income in tax). 3 MARKS
1. $f'(x) = \frac{5}{5x} = \frac{1}{x}$.
2. $f'(x) = \frac{3x^2}{x^3} = \frac{3}{x}$. Or: $\ln(x^3) = 3\ln x$, so derivative = $3/x$.
3. $f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
4. $f'(x) = \frac{(1/x) \cdot x^2 - \ln x \cdot 2x}{x^4} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}$.
5. At $x = 1$: $y = 0$, $m = 1$. Tangent: $y = x - 1$.
1. $\frac{d}{dx}(\ln(kx)) = \frac{k}{kx} = \frac{1}{x}$. The constant $k$ cancels.
2. $\frac{d}{dx}(\ln(x^2+1)) = \frac{2x}{x^2+1} = 0$ when $x = 0$.
3. $\frac{dU}{dW} = \frac{1}{W}$. At $W = 100$: $0.01$. At $W = 10{,}000$: $0.0001$. Marginal utility is 100 times smaller for the wealthier person, meaning each extra dollar provides much less satisfaction.
Q8 (3 marks): Let $u = x^3$, $v = \ln x$ [0.5]. $u' = 3x^2$, $v' = 1/x$ [0.5]. $f'(x) = 3x^2 \ln x + x^3 \cdot \frac{1}{x}$ [0.5] $= 3x^2 \ln x + x^2$ [0.5] $= x^2(3\ln x + 1)$ [1].
Q9 (3 marks): At $x = e$: $y = \ln(e) = 1$ [0.5]. $\frac{dy}{dx} = \frac{1}{x}$ [0.5]. At $x = e$: $m = \frac{1}{e}$ [0.5]. Equation: $y - 1 = \frac{1}{e}(x - e)$ [0.5] $\Rightarrow y = \frac{x}{e} - 1 + 1 = \frac{x}{e}$ [1].
Q10 (3 marks): $\frac{dU}{dW} = \frac{1}{W}$ [0.5]. At $W = 100$: $\frac{dU}{dW} = 0.01$ per dollar [0.25]. This means each additional dollar provides 0.01 units of utility at this wealth level [0.25]. After 10% tax: $W_{new} = 90$ [0.25]. $U_{before} = \ln(100) \approx 4.605$, $U_{after} = \ln(90) \approx 4.500$ [0.5]. Decrease = $4.605 - 4.500 = 0.105$. Percentage decrease = $\frac{0.105}{4.605} \times 100\% \approx 2.28\%$ [0.5]. This supports progressive taxation because the utility loss from a 10% tax is much smaller for a wealthy person (2.28% utility loss) than for a poor person [0.5]. For example, at $W = 10$, a 10% tax reduces utility from $\ln(10) \approx 2.303$ to $\ln(9) \approx 2.197$, a 4.6% loss — roughly double the proportional impact [0.25].
Climb platforms using derivatives of ln x, chain rule, and tangent equations. Pool: lesson 11.
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