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 7 - 2012 - Paper 5

To differentiate $\frac{1 - 10x}{(2x - 1)^5}$, we use the quotient rule, which states if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$.

Let:

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

Finding $g'$ and $h'$:

• $g' = -10$
• Using the chain rule, $h' = 5(2x - 1)^4 \cdot 2 = 10(2x - 1)^4$

Now applying the quotient rule: $f'(x) = \frac{(-10)(2x - 1)^5 - (1 - 10x)(10(2x - 1)^4)}{(2x - 1)^{10}}$
Factoring out $(2x - 1)^4$ gives: $f'(x) = \frac{(2x - 1)^4 [(-10)(2x - 1) - 10(1 - 10x)]}{(2x - 1)^{10}}$
Simplifying: $= \frac{(2x - 1)^4 [-20x + 10 - 10 + 100x]}{(2x - 1)^{10}}$
$= \frac{(2x - 1)^4 (80x)}{(2x - 1)^{10}}$
Thus, the final answer in simplest form is: $\frac{8x}{(2x - 1)^6}$

To find $\frac{dy}{dx}$, we differentiate both sides with respect to $x$.

Starting with $x = 3 \tan(2y)$, we differentiate with implicit differentiation: $\frac{dx}{dx} = 3 \cdot \sec^2(2y) \cdot 2 \frac{dy}{dx}$
This simplifies to: $1 = 6 \sec^2(2y) \frac{dy}{dx}$
Thus: $\frac{dy}{dx} = \frac{1}{6 \sec^2(2y)}$

Now, substituting $y$ in terms of $x$, from the equation: $\tan(2y) = \frac{x}{3}$ implies that :
$2y = \tan^{-1}\left(\frac{x}{3}\right)$
Using the identity $\sec^2(2y) = 1 + \tan^2(2y)$ gives: $\sec^2(2y) = 1 + \left(\frac{x}{3}\right)^2$
Substituting this into our expression for $\frac{dy}{dx}$ we get: $\frac{dy}{dx} = \frac{1}{6\left(1 + \frac{x^2}{9}\right)} = \frac{1}{6 \left(\frac{9 + x^2}{9}\right)}$
Thus, $\frac{dy}{dx} = \frac{9}{6(9 + x^2)} = \frac{3}{2(9 + x^2)}$