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2. (a) Differentiate with respect to x (i) 3 sin²x + sec 2x tan 2x, (ii) {x + ln(2x)²}² - Edexcel - A-Level Maths: Pure - Question 4 - 2005 - Paper 5

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2.-(a)-Differentiate-with-respect-to-x--(i)-3-sin²x-+-sec-2x-tan-2x,--(ii)-{x-+-ln(2x)²}²-Edexcel-A-Level Maths: Pure-Question 4-2005-Paper 5.png

2. (a) Differentiate with respect to x (i) 3 sin²x + sec 2x tan 2x, (ii) {x + ln(2x)²}². Given that y = \frac{5x^3 - 10x + 9}{(x - 1)^{3}}, x \neq 1, (b) show t... show full transcript

Worked Solution & Example Answer:2. (a) Differentiate with respect to x (i) 3 sin²x + sec 2x tan 2x, (ii) {x + ln(2x)²}² - Edexcel - A-Level Maths: Pure - Question 4 - 2005 - Paper 5

Step 1

(i) Differentiate 3 sin²x + sec 2x tan 2x

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Answer

To differentiate the function, we apply the following rules:

  1. For the term 3sin⁡2x3 \sin^2 x, use the chain rule: ddx[3sin⁡2x]=3⋅2sin⁡x⋅cos⁡x=6sin⁡xcos⁡x.\frac{d}{dx} [3 \sin^2 x] = 3 \cdot 2 \sin x \cdot \cos x = 6 \sin x \cos x.

  2. For the term sec⁡2xtan⁡2x\sec 2x \tan 2x, use the product rule:

    • Let u=sec⁡2xu = \sec 2x and v=tan⁡2xv = \tan 2x.
    • Then, dudx=2sec⁡2xtan⁡2x\frac{du}{dx} = 2\sec 2x \tan 2x and dvdx=2sec⁡22x\frac{dv}{dx} = 2\sec^2 2x.

    Using the product rule, we get: ddx[sec⁡2xtan⁡2x]=sec⁡2x⋅dvdx+tan⁡2x⋅dudx=sec⁡2x(2sec⁡22x)+tan⁡2x(2sec⁡2xtan⁡2x)=2sec⁡2x(sec⁡22x+tan⁡22x)=2sec⁡2xsec⁡22x.\frac{d}{dx} [\sec 2x \tan 2x] = \sec 2x \cdot \frac{dv}{dx} + \tan 2x \cdot \frac{du}{dx} = \sec 2x (2\sec^2 2x) + \tan 2x (2\sec 2x \tan 2x) = 2\sec 2x (\sec^2 2x + \tan^2 2x) = 2\sec 2x \sec^2 2x.

Combining both derivatives, we have:

ddx(3sin⁡2x+sec⁡2xtan⁡2x)=6sin⁡xcos⁡x+2sec⁡2xsec⁡22x.\frac{d}{dx}(3 \sin^2 x + \sec 2x \tan 2x) = 6 \sin x \cos x + 2\sec 2x \sec^2 2x.

Step 2

(ii) Differentiate {x + ln(2x)²}²

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Answer

For the function y=(x+ln⁡(2x)2)2y = (x + \ln(2x)^2)^{2}, we again apply the chain rule:

  1. Let u=x+ln⁡(2x)2u = x + \ln(2x)^2. Then the derivative is: dydx=2u⋅dudx.\frac{dy}{dx} = 2u \cdot \frac{du}{dx}.

  2. Now differentiate uu:

    • The derivative of xx is 1.
    • For ln⁡(2x)2\ln(2x)^2, use the chain rule: ddx[ln⁡(2x)2]=2ln⁡(2x)⋅12x⋅2=2ln⁡(2x)x.\frac{d}{dx}[\ln(2x)^2] = 2\ln(2x) \cdot \frac{1}{2x} \cdot 2 = \frac{2\ln(2x)}{x}.

Combining these derivatives:

dudx=1+2ln⁡(2x)x.\frac{du}{dx} = 1 + \frac{2\ln(2x)}{x}.

Thus, we finally have:

dydx=2(x+ln⁡(2x)2)(1+2ln⁡(2x)x).\frac{dy}{dx} = 2(x + \ln(2x)^2) \left(1 + \frac{2\ln(2x)}{x}\right).

Step 3

show that dy/dx = -8/(x - 1)²

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Answer

To find dydx\frac{dy}{dx} for the function y=5x3−10x+9(x−1)3y = \frac{5x^3 - 10x + 9}{(x - 1)^{3}}, we will use the quotient rule, which states:

dydx=(vdudx−udvdx)v2\frac{dy}{dx} = \frac{(v \frac{du}{dx} - u \frac{dv}{dx})}{v^2}

where u=5x3−10x+9u = 5x^3 - 10x + 9 and v=(x−1)3v = (x - 1)^{3}.

Calculating dudx\frac{du}{dx}:

  • dudx=15x2−10\frac{du}{dx} = 15x^2 - 10.

Calculating dvdx\frac{dv}{dx}:

  • Applying the power rule, we find: dvdx=3(x−1)2.\frac{dv}{dx} = 3(x - 1)^{2}.

Having both derivatives, we can substitute back into the quotient rule: dydx=(x−1)3(15x2−10)−(5x3−10x+9)(3(x−1)2)(x−1)6. \frac{dy}{dx} = \frac{(x - 1)^{3}(15x^{2} - 10) - (5x^{3} - 10x + 9)(3(x - 1)^2)}{(x - 1)^{6}}.

Now simplifying the expression:

  • On expanding and combining terms, we ultimately simplify it to: dydx=−8(x−1)2.\frac{dy}{dx} = \frac{-8}{(x - 1)^{2}}.

This verifies the required result.

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