7. (a) By writing sec x as $ \frac{1}{\cos x} $, show that $ \frac{d}{dx}(\sec x) = \sec x \tan x $. (3) Given that $ y = e^{3x} \sec 3x $, (b) find $ \frac{dy}{dx} $. (4) The curve with equation $ y = e^{3x} \sec 3x $, $ -\frac{\pi}{6} < x < \frac{\pi}{6} $, has a minimum turning point at $ (a, b) $. (c) Find the values of the constants $ a $ and $ b $, giving your answers to 3 significant figures. (4)

A-Level Edexcel Maths Pure Functions: 7. (a) By writing sec x as \( \f

7. (a) By writing sec x as $ \frac{1}{\cos x} $, show that $ \frac{d}{dx}(\sec x) = \sec x \tan x $ - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2

Question 9

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7. (a) By writing sec x as $ \frac{1}{\cos x} $, show that $ \frac{d}{dx}(\sec x) = \sec x \tan x $. (3) Given that $ y = e^{3x} \sec 3x $, (b) find \( \frac{... show full transcript

Worked Solution & Example Answer:7. (a) By writing sec x as $ \frac{1}{\cos x} $, show that $ \frac{d}{dx}(\sec x) = \sec x \tan x $ - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2

Step 1

By writing sec x as $ \frac{1}{\cos x} $, show that $ \frac{d}{dx}(\sec x) = \sec x \tan x $

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Answer

To show that ( \frac{d}{dx}(\sec x) = \sec x \tan x ), we start by rewriting secant:

[ \sec x = \frac{1}{\cos x} ]

Using the quotient rule for differentiation, we have:

[ \frac{dy}{dx} = \frac{(-\sin x)(1)}{\cos^2 x} = -\frac{\sin x}{\cos^2 x} ]

Since ( \sec x = \frac{1}{\cos x} ), we can express the derivative as:

[ \frac{dy}{dx} = -\frac{\sin x}{\cos^2 x} = \sec x \tan x ]

Thus, we have shown that ( \frac{d}{dx}(\sec x) = \sec x \tan x ).

Step 2

find $ \frac{dy}{dx} $

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Answer

Given ( y = e^{3x} \sec 3x ), we apply the product rule:

Let: [ u = e^{3x} \quad \text{and} \quad v = \sec 3x ]

Then, [ \frac{du}{dx} = 3e^{3x} \quad \text{and} \quad \frac{dv}{dx} = 3\sec 3x \tan 3x ]

Now, using the product rule ( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} ):

[ \frac{dy}{dx} = e^{3x}(3\sec 3x \tan 3x) + \sec 3x(3e^{3x}) ] [ \frac{dy}{dx} = 3e^{3x} \sec 3x (\tan 3x + 1) ]

Step 3

Find the values of the constants $ a $ and $ b $

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Answer

To find the turning point, we set ( \frac{dy}{dx} = 0 ):

[ 3e^{3x} \sec 3x (\tan 3x + 1) = 0 ]

Since ( e^{3x} ) and ( \sec 3x ) are never zero, we need ( \tan 3x + 1 = 0 ) which gives:

[ \tan 3x = -1 \quad \Rightarrow \quad 3x = -\frac{\pi}{4} + n\pi ]

For ( x = -\frac{\pi}{12} ), we have:

Plugging ( x = -\frac{\pi}{12} ) back into the equation for ( y ): [ y = e^{-\frac{\pi}{4}} \sec(-\frac{\pi}{4}) \Rightarrow \sec(-\frac{\pi}{4}) = \sqrt{2} ]

Thus: [ a = -\frac{\pi}{12}, , b = e^{-\frac{\pi}{4}} \cdot \sqrt{2} \approx 0.812] and rounding gives ( a \approx -0.196 ) and ( b \approx 0.812 ).

7. (a) By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d}{dx}(\sec x) = \sec x \tan x \) - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2

Worked Solution & Example Answer:7. (a) By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d}{dx}(\sec x) = \sec x \tan x \) - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2

By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d}{dx}(\sec x) = \sec x \tan x \)

find \( \frac{dy}{dx} \)

Find the values of the constants \( a \) and \( b \)

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