Differentiation: gradient, stationary points, tangents & velocity

When asked for points with a given gradient, students equated $\dfrac{dy}{dx}$ to 0 instead of to that gradient and lost the method mark.

From $x^2 = 4$ students gave only $x = 2$, ignoring $x = -2$; conversely $\sqrt[3]{8} = 2$ only (not $\pm 2$). Know which sign(s) are valid.

Asked for the equation of the tangent, students stopped at the coordinates $(2, -96)$; subsequent incorrect working also lost the final mark.

In $y = x^2 - 6x + 4$ the $+4$ differentiates to $0$, giving $\dfrac{dy}{dx} = 2x - 6$. Carrying the $4$ through (e.g. writing $2x - 6 + 4$) is a common slip.

Differentiate the whole expression first, then put the $x$-value in. Substituting $x$ into $y$ before differentiating gives a number, which differentiates to 0 and earns nothing.