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5. (a) Differentiate \( \frac{\cos 2x}{\sqrt{x}} \) with respect to x - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 8
Question 6
5. (a) Differentiate \( \frac{\cos 2x}{\sqrt{x}} \) with respect to x. (b) Show that \( \frac{d}{dx}(\sec^2 3x) \) can be written in the form \( \... show full transcript
Worked Solution & Example Answer:5. (a) Differentiate \( \frac{\cos 2x}{\sqrt{x}} \) with respect to x - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 8
Step 1
Differentiate \( \frac{\cos 2x}{\sqrt{x}} \) with respect to x
Answer
To differentiate ( \frac{\cos 2x}{\sqrt{x}} ), we will use the quotient rule:
Let ( u = \cos 2x ) and ( v = \sqrt{x} ).
The quotient rule states:
Calculating ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
- ( \frac{du}{dx} = -2 \sin 2x )
- ( \frac{dv}{dx} = \frac{1}{2\sqrt{x}} )
Now substituting these values into the quotient rule:
This simplifies to:
Step 2
Show that \( \frac{d}{dx}(\sec^2 3x) \) can be written in the form \( \mu(\tan 3x + \tan 3x) \)
Answer
To show that ( \frac{d}{dx}(\sec^2 3x) ) can be written in the desired form, we first differentiate ( \sec^2 3x ) using the chain rule:
We know that: ( \frac{d}{dx}(\sec^2 u) = 2\sec^2 u \tan u \cdot \frac{du}{dx} ).
Letting ( u = 3x ):
( \frac{d}{dx}(\sec^2 3x) = 2\sec^2(3x) \cdot \tan(3x) \cdot 3 = 6\sec^2(3x) \tan(3x) )
Thus, we can express this as: ( \mu(\tan 3x + \tan 3x) ) by letting ( \mu = 6\sec^2(3x) ).
Step 3
Given \( x = 2 \sin(\frac{y}{3}) \), find \( \frac{dy}{dx} \)
Answer
We start with the equation: ( x = 2 \sin(\frac{y}{3}) ).
To find ( \frac{dy}{dx} ), we will use implicit differentiation:
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Differentiating both sides with respect to x: ( \frac{dx}{dx} = 2 \cos(\frac{y}{3}) imes \frac{1}{3} \cdot \frac{dy}{dx} )
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Thus, we have: ( 1 = \frac{2}{3} \cos(\frac{y}{3}) \cdot \frac{dy}{dx} )
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Solving for ( \frac{dy}{dx} ): ( \frac{dy}{dx} = \frac{3}{2 \cos(\frac{y}{3})} )
Finally, substituting back for y yields the answer in terms of x if required.