Differentiation and Integration

Power rule. $\dfrac{d}{dx}(x^n) = nx^{n-1}$. Bring power down, reduce by 1. Works for any real $n$ (including negative and fractional).

Chain rule. For composite functions $y = f(g(x))$: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ where $u = g(x)$.

Example: $y = (3x + 1)^4$. Let $u = 3x + 1$, so $y = u^4$. Then $\frac{dy}{dx} = 4u^3 \cdot 3 = 12(3x + 1)^3$.

Product rule. $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$.

Quotient rule. $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v u' - u v'}{v^2}$.

Trig derivatives (x in radians).

• $(\sin x)' = \cos x$.
• $(\cos x)' = -\sin x$.
• $(\tan x)' = \sec^2 x$.

Exponential and log.

• $(e^x)' = e^x$.
• $(\ln x)' = \frac{1}{x}$.

Cambridge tip. Memorise the four rules and the four 'special' derivatives ($\sin, \cos, e^x, \ln$). Combine them via chain rule for everything else.

Stationary point. Where the GRADIENT is ZERO. Could be a maximum, minimum, or stationary point of inflection.

Procedure.

1. Differentiate to get $\frac{dy}{dx}$.
2. Set $\frac{dy}{dx} = 0$ and solve for $x$.
3. Find the corresponding $y$ values by substituting back.
4. Classify using the second derivative test.

Second derivative test.

• $\dfrac{d^2 y}{dx^2} > 0$: MINIMUM (curve concave up at this point).
• $\dfrac{d^2 y}{dx^2} < 0$: MAXIMUM (curve concave down).
• $\dfrac{d^2 y}{dx^2} = 0$: INCONCLUSIVE — use first-derivative test (sign of slope on either side).

Worked walkthrough. $y = x^3 - 6x^2 + 9x + 1$.

• $\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$. Zeros at $x = 1, 3$.
• $y(1) = 5$, $y(3) = 1$.
• $\frac{d^2 y}{dx^2} = 6x - 12$. At $x = 1$: $-6 < 0$ → MAX. At $x = 3$: $6 > 0$ → MIN.

Optimisation problems. Real-world max/min: model the quantity to be optimised, differentiate, set to zero, classify.

Cambridge tip. Always classify with the second derivative test (or first-derivative test if needed). Mark scheme awards a separate mark for classification.

Power rule for integration. $\int x^n \, dx = \frac{x^{n+1}}{n + 1} + C, \quad n \ne -1.$

Special case. $\int \frac{1}{x} \, dx = \ln|x| + C$.

Indefinite integral. Always include $+C$ (constant of integration).

Definite integral. $\int_a^b f(x) \, dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$. No $+C$ needed (it cancels).

Area under a curve.

• If $y > 0$ over $[a, b]$: area = $\int_a^b y \, dx$.
• If $y < 0$ over $[a, b]$: area = $\left|\int_a^b y \, dx\right|$ (take absolute value).
• If $y$ changes sign: split the integral at zeros, take absolute values, sum.

Trig and exp integrals (reverse the derivatives).

• $\int \cos x \, dx = \sin x + C$.
• $\int \sin x \, dx = -\cos x + C$.
• $\int e^x \, dx = e^x + C$.

Cambridge tip. Mark scheme always penalises missing $+C$ on indefinite integrals. Make it a habit.

For a particle moving in 1D:

• Position $s(t)$.
• Velocity $v = \frac{ds}{dt}$ (derivative of position).
• Acceleration $a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$.

Going backwards.

• $v = \int a \, dt$.
• $s = \int v \, dt$.

Each integration introduces a constant of integration — usually determined by initial conditions ($v(0)$ or $s(0)$).

Worked walkthrough. A particle has $v = 6t^2 + 2t$. Find acceleration and total displacement during $0 \le t \le 2$.

• $a = \frac{dv}{dt} = 12t + 2$. At $t = 2$: $a = 26$.
• $s = \int_0^2 (6t^2 + 2t) \, dt = [2t^3 + t^2]_0^2 = 16 + 4 - 0 = 20$.

Stationary in motion. Velocity = 0 → particle momentarily at rest (e.g. at the highest point of a vertical throw). Often asked: 'Find when the particle is instantaneously at rest.'

Cambridge tip. Always note the initial conditions when integrating to avoid losing constants.