Photo AI
7. (a) By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d(\sec x)}{dx} = \sec x \tan x \) - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Question 9
7. (a) By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d(\sec x)}{dx} = \sec x \tan x \). (b) Given that \( y = e^{x} \sec 3x \), find \( \frac{dy}... show full transcript
Worked Solution & Example Answer:7. (a) By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d(\sec x)}{dx} = \sec x \tan x \) - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Step 1
By writing sec x as \( \frac{1}{\cos x} \), show that \( \frac{d(\sec x)}{dx} = \sec x \tan x \)
Answer
To prove this, we start with the definition of secant:
[ y = \sec x = \frac{1}{\cos x} ]
Differentiating using the quotient rule:
[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{\cos x}\right) = \frac{0 \cdot \cos x - (-\sin x)(1)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} ]
Now, since ( \sec x = \frac{1}{\cos x} ), we can rewrite ( \frac{\sin x}{\cos^2 x} ) as ( \sec x \tan x ):
[ \frac{dy}{dx} = \sec x \tan x ]
Step 2
Given that \( y = e^{x} \sec 3x \), find \( \frac{dy}{dx} \)
Answer
Using the product rule, where:
[ u = e^{x} \quad \text{and} \quad v = \sec 3x ]
We find ( \frac{du}{dx} = e^{x} ) and ( \frac{dv}{dx} = 3 \sec 3x \tan 3x ).
Applying the product rule:
[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = e^{x}(3\sec 3x \tan 3x) + \sec 3x(e^{x}) ]
Thus, we have:
[ \frac{dy}{dx} = e^{x} \sec 3x(3 \tan 3x + 1) ]
Step 3
The curve with equation \( y = e^{x} \sec 3x, \frac{-\pi}{6} < x < \frac{\pi}{6} \), has a minimum turning point at (a, b).
Answer
To find the turning points, set ( \frac{dy}{dx} = 0 ):
[ e^{x} \sec 3x(3 \tan 3x + 1) = 0 ]
Since ( e^{x} \sec 3x \neq 0 ), we solve:
[ 3 \tan 3x + 1 = 0 \qquad \Rightarrow \qquad \tan 3x = -\frac{1}{3} ]
Calculating ( 3x ):
[ 3x = \tan^{-1}\left(-\frac{1}{3}\right) ]
Solving this provides us with specific values for x. We can calculate the corresponding points (a, b) by substituting back into the original equation for y to obtain ( a ) and ( b ) as follows.