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 $\frac{1-10x}{(2x-1)^{5}}$, we will use the quotient rule. Let:

• $f = 1 - 10x$
• $g = (2x - 1)^{5}$

We have:

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

Using the quotient rule:

$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g \frac{df}{dx} - f \frac{dg}{dx}}{g^{2}}$

Substituting into the quotient rule gives:

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

This simplifies to:

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

So the simplified answer becomes:
$\frac{10(-10x - 9)}{(2x - 1)^{6}}$

To find $\frac{dy}{dx}$ given $x = 3 \tan 2y$, we will differentiate implicitly.
Starting with:
$\frac{dx}{dy} = 3 \cdot \frac{d}{dy}(\tan 2y) = 3 \cdot 2 \sec^{2}(2y) \cdot \frac{dy}{dy}$
Thus, we have:
$\frac{dx}{dy} = 6 \sec^{2}(2y)$
To find $\frac{dy}{dx}$, we take the reciprocal:
$\frac{dy}{dx} = \frac{1}{6 \sec^{2}(2y)}$
Now, using the identity $\sec^{2}(2y) = 1 + \tan^{2}(2y)$ and substituting back for $x$:
Since $\tan 2y = \frac{x}{3}$, we have $\sec^{2}(2y) = 1 + \left(\frac{x}{3}\right)^{2} = 1 + \frac{x^{2}}{9}$, therefore:
$\frac{dy}{dx} = \frac{1}{6(1 + \frac{x^{2}}{9})} = \frac{1}{6 + \frac{2}{3}x^{2}}$
This is the expression for $\frac{dy}{dx}$ in terms of $x$.