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A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find \( \frac{dy}{dx} \), giving your answer in the form \( (mx^2 + n) \cos 3x \), where \( m \) and \( n \) are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2
Question 8
A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find \( \frac{dy}{dx} \), giving your answer in the form \( (mx^2 + n) \... show full transcript
Worked Solution & Example Answer:A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find \( \frac{dy}{dx} \), giving your answer in the form \( (mx^2 + n) \cos 3x \), where \( m \) and \( n \) are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2
Step 1
Find \( \frac{dy}{dx} \)
Answer
To find the derivative ( \frac{dy}{dx} ), we will apply the product rule. The given equation is:
[
y = 2x , ext{cos} , 3x + (3x^2 - 4) , ext{sin} , 3x
]
Using the product rule ( \frac{d}{dx}[u \cdot v] = u'v + uv' ), we derive each term:
- For the first term ( 2x , ext{cos} , 3x ):
- Let ( u = 2x ) and ( v = \text{cos} , 3x ):
- ( u' = 2 ) and ( v' = -3 \text{sin} , 3x )
- Applying the product rule:
[ \frac{d}{dx}[2x \cdot \text{cos} , 3x] = 2 \cdot \text{cos} , 3x + 2x \cdot (-3 \text{sin} , 3x) ] [ = 2 \text{cos} , 3x - 6x \text{sin} , 3x ]
- For the second term ( (3x^2 - 4) \text{sin} , 3x ):
- Let ( u = 3x^2 - 4 ) and ( v = \text{sin} , 3x ):
- ( u' = 6x ) and ( v' = 3 \text{cos} , 3x )
- Applying the product rule:
[ \frac{d}{dx}[(3x^2 - 4) \cdot \text{sin} , 3x] = 6x \cdot \text{sin} , 3x + (3x^2 - 4)(3 \text{cos} , 3x) ] [ = 6x \text{sin} , 3x + (3x^2 - 4) imes 3 \text{cos} , 3x ]
Combining both results, we have:
[ \frac{dy}{dx} = (2 - 18x) \text{cos} , 3x + (3x^2 - 4)(3 \text{cos} , 3x) + 6x \text{sin} , 3x ] Combining like terms, the answer can be rearranged into the form ( (mx^2 + n) \cos 3x ).
Step 2
Show that the x-coordinates of the points of inflection satisfy the equation
Answer
To find the points of inflection, we need to set the second derivative ( \frac{d^2y}{dx^2} ) to zero.
First, we calculate ( \frac{dy}{dx} ) from part (a):
[\frac{dy}{dx} = (2 - 18x) \cos 3x + (3x^2 - 4)(3 \cos 3x) + 6x \sin 3x.]
Next, we derive ( \frac{dy}{dx} ) to find ( \frac{d^2y}{dx^2} ) using the product rule again. Once we arrive at the expression for the second derivative, we equate it to zero:
[
\frac{d^2y}{dx^2} = 0]
Solving this equation will lead us to the coordinates of the points of inflection. After simplifying and manipulating the equation, we show that:
[ \cot 3x = \frac{9x^2 - 10}{6x} ]
This confirms the points of inflection satisfy the required equation.