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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 Mechanics - 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) \co... 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 Mechanics - Question 8 - 2017 - Paper 2
Step 1
Find \(\frac{dy}{dx}\)
Answer
To differentiate the given equation, we will use the product rule, which states that if (y = u , v), then (\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.)
Let:
- (u = 2x) and (v = \cos 3x)
- (u = 3x^2 - 4) and (v = \sin 3x)
First term derivative:
- (\frac{dy}{dx} = 2 \cdot \cos 3x + 2x \cdot (-3 \sin 3x) = 2 \cos 3x - 6x \sin 3x.)
Second term derivative: 2. (\frac{dy}{dx} = (3x^2 - 4) \cdot 3 \cos 3x + \sin 3x \cdot 6x = 3(3x^2 - 4) \cos 3x + 6x \sin 3x.)
Combining both terms, we get:
$$\frac{dy}{dx} = (9x^2 - 10) \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 to zero, that is:
- Calculate (\frac{d^2y}{dx^2}).
- Set (\frac{d^2y}{dx^2} = 0) and solve for (x).
Starting from the first derivative (\frac{dy}{dx} = (9x^2 - 10) \cos 3x), we differentiate it again, applying the product rule:
(\frac{d^2y}{dx^2} = (9x^2 - 10)(-3 \sin 3x) + 18x \cos 3x.)
Setting this to zero gives:
From the equation, we can rearrange and simplify to show that: