Calculus (JAMB bridge)

The gradient of the tangent at that point. $\dfrac{dy}{dx}$ gives it at every point.

$\dfrac{dy}{dx} = nax^{n - 1}$: bring the power down, then reduce the power by 1.

$\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$.

$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$: outside first, then times the derivative of the inside. So $\dfrac{d}{dx}\sin 7x = 7\cos 7x$.

For $y = uv$: $\dfrac{dy}{dx} = u'v + uv'$.

For $y = \dfrac uv$: $\dfrac{dy}{dx} = \dfrac{vu' - uv'}{v^2}$.

Solve $\dfrac{dy}{dx} = 0$. Then check: $\dfrac{d^2y}{dx^2} < 0$ is a maximum and $> 0$ a minimum, or look at the gradient's sign either side (+ 0 − is a maximum).

Chain rule: $\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt} = 2\pi r \times \dfrac{dr}{dt}$.

Differentiate: $v = \dfrac{ds}{dt}$ and $a = \dfrac{dv}{dt}$.

$\dfrac{ax^{n + 1}}{n + 1} + c$, for $n \ne -1$: add 1 to the power, divide by the new power, add $c$.

Every curve $y = F(x) + c$ has the same gradient. A given point, put into $y$, picks out $c$.

$\int_a^b f(x)\,dx = F(b) - F(a)$.

Integrate (top − bottom) between the points where they cross.

$V = \pi\int_a^b y^2\,dx$: each slice is a disc of radius $y$.