Jodie is attempting to use differentiation from first principles to prove that the gradient of $y = ext{sin} \, x$ is zero when $x = rac{ ext{pi}}{2}$ - AQA - A-Level Maths Pure - Question 11 - 2019 - Paper 1

To find the gradient of the curve at point A, we start from the gradient of the chord AB formula:

ext{Gradient of chord } AB = rac{ ext{sin} \left(\frac{\text{pi}}{2} + h\right) - ext{sin} \left(\frac{\text{pi}}{2}\right)}{h}

In Step 4, we need to evaluate this as $h$ approaches 0.

We rewrite the limit as:

$$ext{lim}_{h \to 0} \frac{ ext{sin} \left(\frac{\text{pi}}{2} + h\right) - ext{sin} \left(\frac{\text{pi}}{2}\right)}{h}$$

Using the limit properties, we substitute for $\text{sin}\left(\frac{\text{pi}}{2} + h\right)$ which gives:

$$ext{sin}\left(\frac{\text{pi}}{2}\right) \text{cos}(h) + \text{cos}\left(\frac{\text{pi}}{2}\right)\text{sin}(h)$$

Thus we simplify further:

1. As $h \to 0$, $\text{cos}(h)\to 1$ and $\text{sin}(h) \to 0$.
2. Ultimately, we need the expression:

$$ext{sin} \left(\frac{\text{pi}}{2}\right) + 0 = 1$$

Therefore, the gradient of the curve at A is represented as:

$ext{sin} \left(\frac{\text{pi}}{2}\right) \cdot 0 + \text{cos} \left(\frac{\text{pi}}{2}\right) \cdot 1 = 0.$