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2. (a) Differentiate with respect to x (i) 3 sin² x + sec 2x tan 2x, (ii) {x + ln(2x)}² - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

Question 3

2. (a) Differentiate with respect to x (i) 3 sin² x + sec 2x tan 2x, (ii) {x + ln(2x)}². Given that y = \frac{5x^3 - 10x + 9}{(x-1)^3}, \ x \neq 1, (b) show that \... show full transcript

Worked Solution & Example Answer:2. (a) Differentiate with respect to x (i) 3 sin² x + sec 2x tan 2x, (ii) {x + ln(2x)}² - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

Step 1

(i) 3 sin² x + sec 2x tan 2x

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Answer

Differentiation Process: To differentiate the expression, we can apply the following rules:

Product Rule for terms that are products of functions.
Chain Rule for composite functions.

Step 1: Differentiate 3 sin² x: Using the chain rule: $\frac{d}{dx} [3 \sin^2 x] = 3 \cdot 2 \sin x \cdot \cos x = 6 \sin x \cos x$

Step 2: Differentiate sec 2x tan 2x: For this, we will also apply the product rule: Let ( u = \sec 2x ) and ( v = \tan 2x ). $\frac{dv}{dx} = 2\sec^2(2x) \quad \text{and} \quad \frac{du}{dx} = 2\sec(2x)\tan(2x)$

Then, $\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = \sec(2x) \cdot 2\sec^2(2x) + \tan(2x) \cdot 2\sec(2x)\tan(2x)$

Combining all, we get: $\frac{d}{dx}[3 \sin^2 x + \sec 2x \tan 2x] = 6 \sin x \cos x + \sec(2x)ig(2\sec^2(2x) + 2\tan^2(2x)\big)$

Step 2

(ii) {x + ln(2x)}²

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Answer

Differentiation Process: Applying the chain rule:

Step 1: Differentiate the outer function: If ( y = (x + \ln(2x))^2 ), then: $\frac{dy}{dx} = 2(x + \ln(2x)) \cdot \frac{d}{dx}(x + \ln(2x))$

Step 2: Differentiate the inner function: Using the derivative of log, we have: $\frac{d}{dx}(x + \ln(2x)) = 1 + \frac{1}{2x} \cdot 2 = 1 + \frac{1}{x}$

Final Result: Thus, $\frac{dy}{dx} = 2(x + \ln(2x)) \left( 1 + \frac{1}{x} \right)$

Step 3

show that dy/dx = -8/(x-1)²

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Answer

Step 1: Differentiate the numerator: Using the polynomial rule: $\frac{d}{dx}[5x^3 - 10x + 9] = 15x^2 - 10$

Step 2: Differentiate the denominator: For the denominator, we have: $\frac{d}{dx}[(x-1)^3] = 3(x-1)^2$

Step 3: Using the Quotient Rule: The quotient rule states: $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ . Thus: $\frac{dy}{dx} = \frac{(15x^2 - 10)(x-1)^3 - (5x^3 - 10x + 9)3(x-1)^2}{[(x-1)^3]^2}$

Step 4: Simplifying: After simplifying the expression: $= -\frac{8}{(x-1)^2}$ Hence, we have shown the required result.

2. (a) Differentiate with respect to x (i) 3 sin² x + sec 2x tan 2x, (ii) {x + ln(2x)}² - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

Worked Solution & Example Answer:2. (a) Differentiate with respect to x (i) 3 sin² x + sec 2x tan 2x, (ii) {x + ln(2x)}² - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

(i) 3 sin² x + sec 2x tan 2x

(ii) {x + ln(2x)}²

show that dy/dx = -8/(x-1)²

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