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7. Differentiate with respect to $x$, (i) $ rac{1}{x^2} ext{ln}(3x)$ (ii) $ rac{1-10x}{(2x-1)^5$} giving your answer in its simplest form - Edexcel - A-Level Maths: Pure - Question 2 - 2012 - Paper 5

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7.-Differentiate-with-respect-to-$x$,---(i)-$-rac{1}{x^2}--ext{ln}(3x)$---(ii)-$-rac{1-10x}{(2x-1)^5$}-giving-your-answer-in-its-simplest-form-Edexcel-A-Level Maths: Pure-Question 2-2012-Paper 5.png

7. Differentiate with respect to $x$, (i) $ rac{1}{x^2} ext{ln}(3x)$ (ii) $ rac{1-10x}{(2x-1)^5$} giving your answer in its simplest form. (b) Given that $x =... show full transcript

Worked Solution & Example Answer:7. Differentiate with respect to $x$, (i) $ rac{1}{x^2} ext{ln}(3x)$ (ii) $ rac{1-10x}{(2x-1)^5$} giving your answer in its simplest form - Edexcel - A-Level Maths: Pure - Question 2 - 2012 - Paper 5

Step 1

(i) $ rac{1}{x^2} ext{ln}(3x)$

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Answer

To differentiate the function, we will apply the product rule. Let:

  • u = rac{1}{x^2}
  • v=extln(3x)v = ext{ln}(3x)
    Then, the derivative is:

rac{d}{dx} (uv) = u'v + uv'

  1. First, compute u′u':

    u' = rac{d}{dx} (x^{-2}) = -2x^{-3} = - rac{2}{x^3}

  2. Now, compute v′v':

    v' = rac{d}{dx} ( ext{ln}(3x)) = rac{1}{3x} imes 3 = rac{1}{x}

  3. Now substitute into the product rule:

    rac{d}{dx} igg( rac{1}{x^2} ext{ln}(3x)igg) = - rac{2}{x^3} ext{ln}(3x) + rac{1}{x} imes rac{1}{x^2}

    Simplifying gives: - rac{2 ext{ln}(3x)}{x^3} + rac{1}{x^3} = rac{1 - 2 ext{ln}(3x)}{x^3}

Step 2

(ii) $\frac{1-10x}{(2x-1)^5}$

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Answer

Here, we will use the quotient rule. Let:

  • f(x)=1−10xf(x) = 1 - 10x
  • g(x)=(2x−1)5g(x) = (2x - 1)^5

Then, the derivative is:

rac{dy}{dx} = rac{f'g - fg'}{g^2}

  1. Compute f′f':

    f′=−10f' = -10

  2. Compute g′g' using the chain rule:

    g′=5(2x−1)4imes2=10(2x−1)4g' = 5(2x - 1)^4 imes 2 = 10(2x - 1)^4

  3. Substitute f,f′,g,g′f, f', g, g' into the quotient rule:

    rac{dy}{dx} = rac{(-10)(2x - 1)^5 - (1 - 10x)(10(2x - 1)^4)}{(2x - 1)^{10}}

    Simplifying this expression will give: rac{-10(2x - 1) + 10(1 - 10x)}{(2x - 1)^6}

    Simplified, this is: rac{80x - 10}{(2x - 1)^6}.

Step 3

Given that $x = 3 \tan 2y$, find $\frac{dy}{dx}$ in terms of $x$.

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Answer

We start with the equation:

x=3tan⁡2yx = 3 \tan 2y

To find rac{dy}{dx}, we'll differentiate both sides with respect to xx:

  1. Differentiate the left side:

    dxdx=1\frac{dx}{dx} = 1

  2. Differentiate the right side using the chain rule:

    ddx(3tan⁡2y)=3sec⁡2(2y)d(2y)dx=6sec⁡2(2y)dydx\frac{d}{dx}(3 \tan 2y) = 3 \sec^2(2y) \frac{d(2y)}{dx} = 6 \sec^2(2y) \frac{dy}{dx}

Putting it together:

1=6sec⁡2(2y)dydx1 = 6 \sec^2(2y) \frac{dy}{dx}

  1. Thus:

    dydx=16sec⁡2(2y)\frac{dy}{dx} = \frac{1}{6 \sec^2(2y)}

  2. Now, express an(2y) an(2y) in terms of xx:

    From x=3tan⁡(2y)x = 3 \tan(2y), we have: tan⁡(2y)=x3\tan(2y) = \frac{x}{3}

    Therefore, from trigonometric identities, we know that: sec⁡2(2y)=1+tan⁡2(2y)=1+(x3)2=1+x29\sec^2(2y) = 1 + \tan^2(2y) = 1 + \left(\frac{x}{3}\right)^2 = 1 + \frac{x^2}{9}

  3. Substitute back into the expression for rac{dy}{dx}:

    dydx=16(1+x29)=16+23x23=36+23x2\frac{dy}{dx} = \frac{1}{6(1 + \frac{x^2}{9})} = \frac{1}{\frac{6 + \frac{2}{3}x^2}{3}} = \frac{3}{6 + \frac{2}{3}x^2}

In simplest terms: dydx=36+23x2\frac{dy}{dx} = \frac{3}{6 + \frac{2}{3}x^2}

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