Moore's Law predicted that the number of transistors on a microchip doubles approximately every two years. This is exponential growth with base 2: $N = N_0 \cdot 2^{t/2}$. But how fast is this growth at any moment? We cannot use $\frac{d}{dx}(e^x) = e^x$ directly — the base is 2, not $e$. The key insight: any exponential $a^x$ can be rewritten as $e^{x \ln a}$, and then we apply the chain rule. This transforms any base into the natural base, giving us a universal differentiation rule for all exponentials.
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Without calculating, predict: is $\frac{d}{dx}(2^x)$ equal to $2^x$, $2^x \ln(2)$, or $x \cdot 2^{x-1}$? Think about how $2^x$ relates to $e^x$.
Core Content
We rewrite $a^x$ using natural exponentials:
a^x = (e^{ln a})^x = e^{x ln a}
Now differentiate using the chain rule, with $u = x \ln(a)$:
d/dx(a^x) = d/dx(e^{x ln a}) = e^{x ln a} · ln(a) = a^x · ln(a)
This is the general exponential differentiation rule.
When $a = e$: $\ln(e) = 1$, so $\frac{d}{dx}(e^x) = e^x \cdot 1 = e^x$. This confirms our previous result as a special case.
When $a = 2$: $\frac{d}{dx}(2^x) = 2^x \ln(2) \approx 0.693 \cdot 2^x$. The growth is slower than $e^x$.
When $a = 10$: $\frac{d}{dx}(10^x) = 10^x \ln(10) \approx 2.303 \cdot 10^x$. The growth is faster than $e^x$.
For composite functions:
d/dx(a^{f(x)}) = a^{f(x)} · ln(a) · f'(x)
Examples:
Find the derivative of $f(x) = 3^{2x} + x^2$ and evaluate $f'(0)$.
$f'(x)$ and $f'(0)$.
f(x) = 3^{2x} + x^2
For 3^{2x}: a = 3, k = 2
d/dx(3^{2x}) = 3^{2x} · ln(3) · 2 = 2ln(3)·3^{2x}
d/dx(x^2) = 2x
f'(x) = 2ln(3)·3^{2x} + 2x
f'(0) = 2ln(3)·3^0 + 0 = 2ln(3) ≈ 2.197
$f'(x) = 2\ln(3) \cdot 3^{2x} + 2x$ and $f'(0) = 2\ln(3) \approx 2.20$.
Find $\frac{d}{dx}(x \cdot 2^x)$.
Answer:
Product rule: $u = x$, $v = 2^x$. $u' = 1$, $v' = 2^x \ln(2)$. $\frac{d}{dx}(x \cdot 2^x) = 1 \cdot 2^x + x \cdot 2^x \ln(2) = 2^x(1 + x\ln(2))$.
At $x = 0$, all exponentials $a^x$ pass through $(0, 1)$. But their gradients differ:
| Function | Derivative | Gradient at $x = 0$ |
|---|---|---|
| $2^x$ | $2^x \ln(2)$ | $\ln(2) \approx 0.693$ |
| $e^x$ | $e^x$ | $1$ |
| $3^x$ | $3^x \ln(3)$ | $\ln(3) \approx 1.099$ |
| $10^x$ | $10^x \ln(10)$ | $\ln(10) \approx 2.303$ |
The gradient at $x = 0$ is exactly $\ln(a)$. This is why $e$ is special: it is the only base where the gradient at $x = 0$ equals 1.
$\frac{d}{dx}(a^x) = a^x \ln(a)$
$\frac{d}{dx}(a^{kx}) = k a^{kx} \ln(a)$
$a^x = e^{x \ln a}$
$a = e \Rightarrow \ln(e) = 1$
$\frac{d}{dx}(2^x) = 2^x \ln(2)$. It is not simply $2^x$ (that only works for base $e$), and it is not $x \cdot 2^{x-1}$ (that is the power rule, which applies to $x^n$, not $a^x$). The correct answer comes from rewriting $2^x = e^{x \ln 2}$ and applying the chain rule. The factor $\ln(2) \approx 0.693$ scales the derivative down from what it would be if the base were $e$.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Differentiate $f(x) = 4^{2x} + e^{3x}$. Show all working. 3 MARKS
9. Find the gradient of $y = 5^x$ at $x = 1$, giving your answer in exact form. 3 MARKS
10. The number of bacteria in a culture is modelled by $N = 100 \cdot 3^{t/2}$, where $t$ is in hours. Find $\frac{dN}{dt}$ and evaluate it at $t = 4$. Explain what this value means in the context of bacterial growth. Then explain why the derivative $\frac{dN}{dt}$ is proportional to $N$ itself, and identify the constant of proportionality. 3 MARKS
1. $f'(x) = 5^x \ln(5)$.
2. $f'(x) = 2^{3x} \cdot \ln(2) \cdot 3 = 3\ln(2) \cdot 2^{3x}$.
3. $f'(x) = 3^x + x \cdot 3^x \ln(3) = 3^x(1 + x\ln(3))$.
4. $\frac{dy}{dx} = 10^x \ln(10)$. At $x = 0$: $\ln(10) \approx 2.303$.
1. $a^x = e^{x \ln a}$, so $\frac{d}{dx}(a^x) = e^{x \ln a} \cdot \ln(a) = a^x \ln(a)$.
2. $\frac{dN}{dt} = N_0 \cdot 2^{t/2} \cdot \ln(2) \cdot \frac{1}{2} = \frac{\ln(2)}{2} \cdot N$. At $t = 0$: $\frac{dN}{dt} = \frac{\ln(2)}{2} \cdot N_0 \approx 0.347 N_0$. The initial growth rate is about 34.7% of the starting population per unit time.
3. When $a = e$, $\ln(a) = \ln(e) = 1$, so $\frac{d}{dx}(e^{kx}) = k e^{kx} \cdot 1 = k e^{kx}$.
Q8 (3 marks): $\frac{d}{dx}(4^{2x}) = 4^{2x} \cdot \ln(4) \cdot 2 = 2\ln(4) \cdot 4^{2x}$ [1.5]. $\frac{d}{dx}(e^{3x}) = 3e^{3x}$ [1]. $f'(x) = 2\ln(4) \cdot 4^{2x} + 3e^{3x}$ [0.5].
Q9 (3 marks): $\frac{dy}{dx} = 5^x \ln(5)$ [1.5]. At $x = 1$: $\frac{dy}{dx} = 5^1 \cdot \ln(5) = 5\ln(5)$ [1.5].
Q10 (3 marks): $\frac{dN}{dt} = 100 \cdot 3^{t/2} \cdot \ln(3) \cdot \frac{1}{2} = 50\ln(3) \cdot 3^{t/2}$ [0.5]. At $t = 4$: $N = 100 \cdot 3^2 = 900$ bacteria [0.25]. $\frac{dN}{dt} = 50\ln(3) \cdot 3^2 = 450\ln(3) \approx 494$ bacteria per hour [0.5]. This means at $t = 4$ hours, the population is increasing at approximately 494 bacteria per hour [0.25]. $\frac{dN}{dt} = 50\ln(3) \cdot 3^{t/2} = \frac{\ln(3)}{2} \cdot (100 \cdot 3^{t/2}) = \frac{\ln(3)}{2} \cdot N$ [0.5]. So $\frac{dN}{dt}$ is proportional to $N$ with constant of proportionality $\frac{\ln(3)}{2} \approx 0.549$ [0.5]. This means the growth rate is always about 54.9% of the current population size per hour [0.5].
Climb platforms using derivatives of a^x, chain rule, and growth rate comparisons. Pool: lesson 10.
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