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 \[ \mu(\tan 3x + \tan 3x) \] where $ \mu $ is a constant. (c) Given $ x = 2 \sin \left( \frac{y}{3} \right) $, find $ \frac{dy}{dx} $ in terms of x, simplifying your answer.

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5. (a) Differentiate \[ \frac{\cos 2x}{\sqrt{x}} \] with respect to x - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Question 7

$5.-(a)-Differentiate----\[-\frac{\cos-2x}{\sqrt{x}}-\]----with-respect-to-x-Edexcel-A-Level Maths Pure-Question 7-2013-Paper 8.png$

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 7 - 2013 - Paper 8

Step 1

Differentiate $ \frac{\cos 2x}{\sqrt{x}} $ with respect to x.

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Answer

To differentiate ( \frac{\cos 2x}{\sqrt{x}} ), we will use the quotient rule, which states that if you have a function ( \frac{u}{v} ), the derivative is given by ( \frac{u'v - uv'}{v^2} ).

Let ( u = \cos 2x ) and ( v = \sqrt{x} ).

First, we find the derivatives:

( u' = -2\sin 2x ) (using the chain rule)
( v' = \frac{1}{2\sqrt{x}} ) (power rule)

Now applying the quotient rule:

[ \frac{d}{dx} \left( \frac{\cos 2x}{\sqrt{x}} \right) = \frac{(-2\sin 2x)\sqrt{x} - \cos 2x \left(\frac{1}{2\sqrt{x}}\right)}{x} = \frac{-2\sin 2x\sqrt{x} - \frac{\cos 2x}{2\sqrt{x}}}{x} ]

Step 2

Show that $ \frac{d}{dx}(\sec^2 3x) $ can be written in the form $ \mu(\tan 3x + \tan 3x) $.

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Answer

To differentiate ( \sec^2 3x ), we will use the chain rule:

[ \frac{d}{dx}(\sec^2 3x) = 2\sec^2 3x \cdot \tan 3x \cdot 3 = 6\sec^2 3x \tan 3x ]

Now we can express ( 6 \sec^2 3x \tan 3x ) in the required form:

Let ( \mu = 6 ) and thus we can write it as: [ \frac{d}{dx}(\sec^2 3x) = \mu(\tan 3x + \tan 3x) = 6 \cdot 2 \tan 3x = 6 \sec^2 3x \tan 3x ]

Step 3

Given $ x = 2 \sin \left( \frac{y}{3} \right) $, find $ \frac{dy}{dx} $ in terms of x, simplifying your answer.

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Answer

To find ( \frac{dy}{dx} ), we will first differentiate ( x = 2 \sin \left( \frac{y}{3} \right) ) implicitly:

[ \frac{dx}{dy} = 2 \cdot \frac{1}{3} \cos \left( \frac{y}{3} \right) \cdot \frac{dy}{dx} ]

This simplifies to: [ \frac{dx}{dy} = \frac{2}{3} \cos \left( \frac{y}{3} \right) \frac{dy}{dx} ]

Rearranging gives: [ \frac{dy}{dx} = \frac{3}{2 \cos \left( \frac{y}{3} \right)} ]

Now substituting for ( y ) in terms of ( x ): since ( y = 3 \arcsin \left( \frac{x}{2} \right) ), we can find ( \cos \left( \frac{y}{3} \right) = \cos \left( \arcsin \left( rac{x}{2} \right) \right) = \sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{\sqrt{4 - x^2}}{2} )

Thus we have: [ \frac{dy}{dx} = \frac{3}{2\left( \frac{\sqrt{4 - x^2}}{2} \right)} = \frac{3}{\sqrt{4 - x^2}} ]

5. (a) Differentiate \[ \frac{\cos 2x}{\sqrt{x}} \] with respect to x - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Worked Solution & Example Answer:5. (a) Differentiate \[ \frac{\cos 2x}{\sqrt{x}} \] with respect to x - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Differentiate \( \frac{\cos 2x}{\sqrt{x}} \) with respect to x.

Show that \( \frac{d}{dx}(\sec^2 3x) \) can be written in the form \( \mu(\tan 3x + \tan 3x) \).

Given \( x = 2 \sin \left( \frac{y}{3} \right) \), find \( \frac{dy}{dx} \) in terms of x, simplifying your answer.

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