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

To find the gradient of the curve at point A, we start with the limit definition of the derivative.
Firstly, we need to replace $h = 0$. As we take the limit, we set:
$\lim_{h \to 0} \frac{\text{sin}(\frac{\pi}{2}+h) - \text{sin}(\frac{\pi}{2})}{h}$

Now, applying the angle addition formula for sine gives us:
$= \text{sin}(\frac{\pi}{2}) \cos(h) + \text{cos}(\frac{\pi}{2}) \text{sin}(h) - \text{sin}(\frac{\pi}{2})$
Since $\text{sin}(\frac{\pi}{2}) = 1$ and $\text{cos}(\frac{\pi}{2}) = 0$, this simplifies to:
$= (1 \cdot \cos(h)) + (0 \cdot \text{sin}(h)) - 1 = \cos(h) - 1$
So, we rewrite the limit as:
$\lim_{h \to 0} \frac{\cos(h) - 1}{h}$
Using the definition of derivatives, we know that as $h \to 0$, $\cos(h) \to 1$ and $\text{sin}(h) \to h$.
Thus, applying L'Hôpital's Rule:
$= \lim_{h \to 0} \frac{-\sin(h)}{1} = 0$
This confirms that the gradient at point A is indeed $0$.