Chemists use the pH scale, defined as $\text{pH} = -\log_{10}[H^+]$, to measure acidity. When studying how pH changes as a buffer solution is diluted, they need $\frac{d}{dx}(\log_{10} x)$. Biologists studying allometric scaling — how metabolic rate scales with body mass — use logarithms with various bases. The general rule for differentiating $\log_a x$ connects all these applications through a single elegant formula involving the natural logarithm of the base.
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We know $\frac{d}{dx}(\ln x) = \frac{1}{x}$. Predict what $\frac{d}{dx}(\log_{10} x)$ might be. How does the base affect the derivative?
Core Content
Use the change of base formula:
log_a(x) = ln(x) / ln(a)
Since $\ln(a)$ is a constant:
d/dx(log_a(x)) = d/dx(ln(x)/ln(a)) = (1/ln(a)) · d/dx(ln(x)) = (1/ln(a)) · (1/x) = 1/(x ln(a))
For composite functions:
d/dx(log_a(u)) = u'/(u ln(a))
Examples:
At any given $x$, the derivative of $\log_a x$ depends on $\frac{1}{\ln a}$:
| Function | Derivative | Factor $\frac{1}{\ln a}$ |
|---|---|---|
| $\log_2 x$ | $\frac{1.443}{x}$ | $1.443$ |
| $\ln x$ | $\frac{1}{x}$ | $1$ |
| $\log_{10} x$ | $\frac{0.434}{x}$ | $0.434$ |
Larger bases give smaller derivatives because the logarithm grows more slowly. The natural logarithm ($\ln x$) has the "cleanest" derivative with no extra factor.
Find the derivative of $f(x) = \log_3(x^2 + 4)$.
$f'(x)$ in simplified form.
Use chain rule: u = x^2 + 4, so u' = 2x
f'(x) = u'/(u ln(3)) = 2x/((x^2 + 4)ln(3))
$f'(x) = \frac{2x}{(x^2 + 4)\ln(3)}$.
Find $\frac{d}{dx}(x \log_2 x)$.
Answer:
Product rule: $1 \cdot \log_2 x + x \cdot \frac{1}{x \ln(2)} = \log_2 x + \frac{1}{\ln(2)}$.
$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
$\frac{d}{dx}(\log_a u) = \frac{u'}{u \ln a}$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\log_a x = \frac{\ln x}{\ln a}$
$\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln(10)} \approx \frac{0.434}{x}$. The base appears in the denominator as $\ln(a)$ because we convert to natural logs: $\log_a x = \frac{\ln x}{\ln a}$. The derivative is smaller than $\frac{1}{x}$ because $\log_{10} x$ grows more slowly than $\ln x$ (since $10 > e$). Larger bases give smaller derivatives.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Differentiate $f(x) = \log_2(x^2 + 3)$. Show all working. 3 MARKS
9. Find the gradient of $y = \log_{10} x$ at $x = 10$, giving your answer as an exact fraction. 3 MARKS
10. The pH of a solution is given by $\text{pH} = -\log_{10}[H^+]$, where $[H^+]$ is the hydrogen ion concentration in mol/L. Find $\frac{d(\text{pH})}{d[H^+]}$ and evaluate it when $[H^+] = 10^{-7}$ (neutral water). Explain what this value means in terms of how sensitive pH is to changes in concentration at this point. Compare this sensitivity to when $[H^+] = 10^{-2}$ (acidic). 3 MARKS
1. $f'(x) = \frac{1}{x \ln(10)}$.
2. $f'(x) = \frac{2}{2x \ln(5)} = \frac{1}{x \ln(5)}$. Or use $\log_5(2x) = \log_5(2) + \log_5(x)$, so $f'(x) = 0 + \frac{1}{x \ln(5)} = \frac{1}{x \ln(5)}$.
3. $f'(x) = \log_3 x + x \cdot \frac{1}{x \ln(3)} = \log_3 x + \frac{1}{\ln(3)}$.
4. $\frac{dy}{dx} = \frac{1}{x \ln(2)}$. At $x = 8$: $\frac{1}{8 \ln(2)} \approx \frac{1}{5.545} \approx 0.180$.
1. $\log_a x = \frac{\ln x}{\ln a}$, so $\frac{d}{dx}(\log_a x) = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$.
2. $\frac{d(\text{pH})}{d[H^+]} = -\frac{1}{[H^+] \ln(10)}$. At $[H^+] = 10^{-7}$: $-\frac{1}{10^{-7} \ln(10)} \approx -4.34 \times 10^6$ pH units per mol/L. A tiny change in concentration causes a huge pH change.
3. $\frac{d}{dx}(\ln x) = \frac{1}{x}$ has no extra constants. $\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln(10)}$ requires the factor $\frac{1}{\ln(10)}$.
Q8 (3 marks): Let $u = x^2 + 3$, so $u' = 2x$ [0.5]. $f'(x) = \frac{2x}{(x^2 + 3)\ln(2)}$ [2.5].
Q9 (3 marks): $\frac{dy}{dx} = \frac{1}{x \ln(10)}$ [1.5]. At $x = 10$: $\frac{1}{10 \ln(10)}$ [1.5].
Q10 (3 marks): $\frac{d(\text{pH})}{d[H^+]} = -\frac{1}{[H^+] \ln(10)}$ [0.5]. At $[H^+] = 10^{-7}$: $-\frac{1}{10^{-7} \ln(10)} = -\frac{10^7}{\ln(10)} \approx -4.34 \times 10^6$ pH units per mol/L [0.5]. This means a change in concentration of $10^{-9}$ mol/L (one part in 100 of the current concentration) would change pH by about $-4.34 \times 10^6 \times 10^{-9} \approx -0.0043$ units [0.5]. At $[H^+] = 10^{-2}$: $-\frac{1}{10^{-2} \ln(10)} = -\frac{100}{\ln(10)} \approx -43.4$ pH units per mol/L [0.5]. The sensitivity at neutral pH is about 100,000 times greater than at pH 2 [0.5]. This is why the body tightly regulates $[H^+]$ near $10^{-7}$ — even tiny absolute changes in concentration cause large pH swings that could be lethal [0.5].
Climb platforms using derivatives of log_a x, pH sensitivity, and chain rule applications. Pool: lesson 12.
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