The gradient of a curve

$P(2, 8)$ is on $y = x^3$, and $Q$ is the point on the curve where $x = 2 + h$. Find the gradient of the chord $PQ$ in terms of $h$. Hence find the gradient of the curve at $P$.

Giveaway words: chord, $x = 2 + h$, in terms of $h$, hence the gradient at $P$.

A chord is a straight line joining two points on a curve.

The gradient of the curve at $P$ is the gradient of the tangent there: the line that touches the curve at $P$ and runs in the same direction as the curve there.

Slide $Q$ towards $P$ (make $h$ smaller and smaller) and the chord swings round onto the tangent. So the chord gradients close in on the gradient at $P$.

Here $h$ is a small number: the gap in $x$ between $P$ and $Q$.

Same four steps every time:

1. Write $Q$: put its $x$-value (like $2 + h$) into the curve's equation to get its $y$-value.
2. Chord gradient: $\dfrac{\text{rise}}{\text{run}}$. The run from $P$ to $Q$ is $(2 + h) - 2 = h$.
3. Cancel the $h$: expand the top, then divide every term by $h$. This is the step people skip.
4. Let $h \to 0$ (read "$h$ tends to 0": $h$ gets closer and closer to 0). Every term with an $h$ in it vanishes. The number left is the gradient of the curve at $P$.