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

Here, we will use the quotient rule. Let:

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

Then, the derivative is:

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

1. Compute $f'$:

   $f' = -10$

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

   $g' = 5(2x - 1)^4 imes 2 = 10(2x - 1)^4$

3. Substitute $f, f', g, g'$ 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}.

We start with the equation:

$x = 3 \tan 2y$

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

1. Differentiate the left side:

   $\frac{dx}{dx} = 1$

2. Differentiate the right side using the chain rule:

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

Putting it together:

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

3. Thus:

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

4. Now, express $an(2y)$ in terms of $x$:

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

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

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

   $\frac{dy}{dx} = \frac{1}{6(1 + \frac{x^2}{9})} = \frac{1}{\frac{6 + \frac{2}{3}x^2}{3}} = \frac{3}{6 + \frac{2}{3}x^2}$

In simplest terms: $\frac{dy}{dx} = \frac{3}{6 + \frac{2}{3}x^2}$