7. (a) Differentiate with respect to $x$, (i) $\frac{1}{x^2} \ln(3x)$ (ii) $\frac{1 - 10x}{(2x - 1)^5$} giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 5

To differentiate this function, we use the quotient rule. Let:

• $u = 1 - 10x$
• $v = (2x - 1)^5$

Then the quotient rule states:

$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$

1. Calculate $u'$:

$u' = -10$

2. Now, use the chain rule to find $v'$:

$v' = 5(2x - 1)^4 \cdot 2 = 10(2x - 1)^4$

3. Plugging these into the quotient rule gives:

$\frac{dy}{dx} = \frac{(-10)(2x - 1)^5 - (1-10x)(10(2x - 1)^4)}{(2x - 1)^{10}}$

4. Simplifying, we get:

$\frac{dy}{dx} = \frac{-10(2x - 1)^5 + 10(1 - 10x)(2x - 1)^4}{(2x - 1)^{10}}$

5. Factoring gives:

$\frac{dy}{dx} = \frac{10(2x - 1)^4(-x + 1)}{(2x - 1)^{10}}$

Thus:

$\frac{dy}{dx} = \frac{10(1-x)}{(2x - 1)^6}$

To find $\frac{dy}{dx}$, we need to apply implicit differentiation:

1. Differentiate both sides with respect to $x$:

$\frac{dx}{dx} = 3 \sec^2(2y) \cdot \frac{d(2y)}{dx}$

2. Recall that $\frac{d(2y)}{dx} = 2\frac{dy}{dx}$, therefore:

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

3. From this, we can solve for $\frac{dy}{dx}$:

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

4. To express this in terms of $x$, use the original substitution:

Given $x = 3 \tan 2y$, we can find $\tan 2y$ as:

$\tan 2y = \frac{x}{3}$

5. Thus, $\sec^2(2y) = 1 + \tan^2(2y) = 1 + \left(\frac{x}{3}\right)^2 = \frac{9 + x^2}{9}$.

6. Substitute this into the derivative:

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

Finally, we conclude that:

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