What is the only function that is equal to its own rate of change? If a population grows at a rate proportional to its current size, its growth curve is exponential. If a radioactive substance decays at a rate proportional to its remaining mass, its decay curve is exponential. In both cases, the derivative equals the function itself. This is the defining property of $e^x$ — and it makes $e$ the most important number in calculus.
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If $f(x) = e^x$, what might $f'(x)$ be? Think about what "rate of change equals current value" means graphically.
Core Content
From first principles:
f'(x) = lim_{h→0} (e^{x+h} - e^x)/h = lim_{h→0} e^x(e^h - 1)/h = e^x · lim_{h→0} (e^h - 1)/h
The number $e$ is defined so that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. Therefore:
f'(x) = e^x · 1 = e^x
This is extraordinary. No other function (except $f(x) = 0$) equals its own derivative. It means:
Show y = e^x in the first and second quadrants. At x=0, draw a tangent line with slope 1 (45° angle). At x=1, draw a steeper tangent with slope e≈2.7. At x=2, draw an even steeper tangent with slope e²≈7.4. Label each point and its corresponding slope value. Add a small right triangle at each tangent showing rise/run.
If $y = e^u$ where $u$ is a function of $x$, then by the chain rule:
dy/dx = e^u · du/dx
Special case: $y = e^{kx}$ gives $\frac{dy}{dx} = k e^{kx}$. The constant $k$ appears as a multiplier.
Examples:
Notice that the derivative always contains the original exponential multiplied by the derivative of the exponent.
Find the derivative of $f(x) = x^2 e^{3x}$.
$f'(x)$ in simplified form.
Use product rule: u = x^2, v = e^{3x}
u' = 2x
v' = 3e^{3x}
f'(x) = u'v + uv' = 2x·e^{3x} + x^2·3e^{3x}
= e^{3x}(2x + 3x^2)
= xe^{3x}(2 + 3x)
$f'(x) = xe^{3x}(3x + 2)$.
Find the derivative of $f(x) = \frac{e^{2x}}{x + 1}$.
Answer:
$f'(x) = \frac{2e^{2x}(x+1) - e^{2x}(1)}{(x+1)^2} = \frac{e^{2x}(2x + 2 - 1)}{(x+1)^2} = \frac{e^{2x}(2x + 1)}{(x+1)^2}$.
To find the equation of a tangent to $y = e^{f(x)}$ at $x = a$:
Example: Tangent to $y = e^{2x}$ at $x = 0$:
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(e^{kx}) = ke^{kx}$
$\frac{d}{dx}(e^u) = e^u \cdot u'$
Gradient of $y=e^x$ equals $y$-value
$f'(x) = e^x$. The derivative of $e^x$ is $e^x$ itself — this unique property is why $e$ is called the "natural" base. At every point on the curve $y = e^x$, the gradient equals the height. This means the curve grows faster than any polynomial, and its rate of growth is always equal to its current size.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Differentiate $f(x) = (2x + 1)e^{3x}$. Show all working. 3 MARKS
9. Find the equation of the tangent to $y = e^{2x}$ at the point where $x = 0$. Give your answer in the form $y = mx + c$. 3 MARKS
10. A cup of coffee cools according to $T = 20 + 70e^{-0.05t}$, where $T$ is temperature in °C and $t$ is time in minutes. Find $\frac{dT}{dt}$ and evaluate it at $t = 0$ and $t = 20$. Explain what these values mean in the context of cooling coffee, and discuss why the magnitude of the derivative decreases over time. 3 MARKS
1. $f'(x) = 5e^{5x}$.
2. $f'(x) = -3e^{-3x}$.
3. $f'(x) = e^{2x} + x \cdot 2e^{2x} = e^{2x}(1 + 2x)$.
4. At $x = 1$: $y = e$, $m = e$. Tangent: $y - e = e(x - 1) \Rightarrow y = ex$.
1. From first principles: $f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x$. The limit equals 1 by definition of $e$.
2. $f'(x) = -0.1e^{-0.1x}$. The negative means the quantity is decreasing (decay) and the rate is proportional to the current amount.
3. $\frac{d}{dx}(e^{x^2+3x}) = (2x+3)e^{x^2+3x} = 0$ when $2x + 3 = 0$, so $x = -\frac{3}{2}$.
Q8 (3 marks): Let $u = 2x+1$, $v = e^{3x}$ [0.5]. $u' = 2$, $v' = 3e^{3x}$ [0.5]. $f'(x) = 2e^{3x} + (2x+1) \cdot 3e^{3x}$ [0.5] $= e^{3x}(2 + 6x + 3)$ [0.5] $= e^{3x}(6x + 5)$ [1].
Q9 (3 marks): At $x = 0$: $y = e^0 = 1$ [0.5]. $\frac{dy}{dx} = 2e^{2x}$ [0.5]. At $x = 0$: $m = 2e^0 = 2$ [0.5]. Using $y - 1 = 2(x - 0)$ [0.5]: $y = 2x + 1$ [1].
Q10 (3 marks): $\frac{dT}{dt} = 70 \cdot (-0.05)e^{-0.05t} = -3.5e^{-0.05t}$ [0.5]. At $t = 0$: $\frac{dT}{dt} = -3.5$ °C/min [0.5] — the coffee is cooling at 3.5°C per minute initially [0.25]. At $t = 20$: $\frac{dT}{dt} = -3.5e^{-1} \approx -3.5 \times 0.368 \approx -1.29$ °C/min [0.5] — after 20 minutes, it cools much more slowly at about 1.3°C per minute [0.25]. The magnitude decreases because as the coffee approaches room temperature (20°C), the temperature difference driving the cooling becomes smaller [0.5]. This is Newton's Law of Cooling: the rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature [0.5].
Climb platforms using derivatives of e^x, chain rule, and tangent equations. Pool: lesson 9.
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