A lens designer models a curve as y = √x, with x and y in centimetres (cm)…

Step 1. Rationalise. $\dfrac{\sqrt{x + h} - \sqrt{x}}{h} = \dfrac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \dfrac{1}{\sqrt{x + h} + \sqrt{x}}$.

Step 2. The limit. As $h \to 0$ this tends to $\dfrac{1}{2\sqrt{x}}$. So $f'(x) = \dfrac{1}{2\sqrt{x}}$.

Step 3. The slopes. At $x = 4$: $\dfrac{1}{2 \times 2} = 0.25$. At $x = 9$: $\dfrac{1}{2 \times 3} = \dfrac{1}{6} \approx 0.1667$.

Step 4. The tangent at $x = 4$. The point is $(4, 2)$, so $y - 2 = 0.25(x - 4)$, which is $y = 0.25x + 1$. It meets the $y$-axis at $1\ \text{cm}$.