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. Her working is shown below. Step 1 Gradient of chord $AB = \text{sin}(\frac{\pi}{2} + h) - \text{sin}(\frac{\pi}{2})$ Step 2 = $\text{sin}(\frac{\pi}{2}) \cos(h) + \text{cos}(\frac{\pi}{2}) \text{sin}(h)$ Step 3 = $\text{sin}(\frac{\pi}{2}) \cos(h) - 0 + \text{cos}(\frac{\pi}{2}) \text{sin}(h)$ Step 4 For gradient of curve at A, let $h = 0$ then $\text{cos}(h) - 1 = 0$ and $\frac{\text{sin}(h)}{h} \to 1$ Step 5 Hence the gradient of the curve at A is given by $\text{sin}(\frac{\pi}{2}) \cdot 0 + \text{cos}(\frac{\pi}{2}) \cdot 1 = 0$

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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 Pure - Question 11 - 2019 - Paper 1

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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 t... show full transcript

Worked Solution & Example Answer: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 Pure - Question 11 - 2019 - Paper 1

Step 1

For gradient of curve at A, let $h = 0$ then

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Answer

As we move towards the limit where $h$ approaches 0, we simplify the terms.

By the definition of the derivative, we know that: $\lim_{h \to 0} \frac{\text{sin}(h)}{h} = 1$

Thus, substituting in the values as $h$ approaches 0 gives us:

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

Therefore:

The gradient of the curve at $A$ now becomes: $\text{sin}(\frac{\pi}{2}) \cdot 0 + \text{cos}(\frac{\pi}{2}) \cdot 1 = 1 \cdot 0 + 0 \cdot 1 = 0.$

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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 Pure - Question 11 - 2019 - Paper 1

Worked Solution & Example Answer: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 Pure - Question 11 - 2019 - Paper 1

For gradient of curve at A, let $h = 0$ then

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