Jodie is attempting to use differentiation from first principles to prove that the gradient of $y = ext{sin} \, x$ is zero when $x = \frac{\pi}{2}.$ Jodie's teacher tells her that she has made mistakes starting in Step 4 of her working - AQA - A-Level Maths Pure - Question 11 - 2019 - Paper 1

To find the gradient of the curve at point A, we start by applying the limit definition of the derivative. We first replace $h$ with 0:

$\lim_{h \to 0} \frac{\text{sin}(\frac{\pi}{2} + h) - \text{sin}(\frac{\pi}{2})}{h}$

Using the sine addition formula, we simplify this to:

$\frac{\text{sin}(\frac{\pi}{2}) \cos(h) + \text{cos}(\frac{\pi}{2}) \text{sin}(h) - \text{sin}(\frac{\pi}{2})}{h}$

Now substituting $\text{sin}(\frac{\pi}{2}) = 1$ and $\text{cos}(\frac{\pi}{2}) = 0$, we get:

$\lim_{h \to 0} \frac{1 \cdot \cos(h) - 1}{h} + 0$

Next, we evaluate the limits. We know that as $h$ approaches zero, $\cos(h)$ approaches 1, so we focus on:

$\text{lim}_{h \to 0} \frac{\cos(h) - 1}{h} \to 0\text{ and } \frac{\text{sin}(h)}{h} \to 1.$

Thus, the corresponding gradient yields zero:

Hence,
$\text{sin}(\frac{\pi}{2}) \cdot 0 + \text{cos}(\frac{\pi}{2}) \cdot 1 = 0.$