4. (i) Given that $x = sec^2 2y$, $0 < y < \frac{\pi}{4}$ show that $$\frac{dy}{dx} = \frac{1}{4x(x - 1)}$$ (ii) Given that y = $(x^2 + x) \ln 2x$ find the exact value of $$\frac{dy}{dx}$$ at $x = \frac{e}{2}$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 6

We start by applying the product rule:

$\frac{dy}{dx} = (u \cdot v' + v \cdot u')$

where $u = (x^2 + x)$ and $v = \ln(2x)$.

Calculating $u'$:

$u' = 2x + 1$

Calculating $v'$ using the chain rule:

$v' = \frac{1}{2x} \cdot 2 = \frac{1}{x}$

Thus combining these:

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

Now substituting $x = \frac{e}{2}$:

Calculate $u = \left(\frac{e^2}{4} + \frac{e}{2}\right)$ and $v = \ln(e) = 1$.

Putting these values into the expression for $\frac{dy}{dx}$ yields:

$\frac{dy}{dx} = 0 + 1(2 \cdot \frac{e}{2} + 1) = e + 1$

To differentiate $f(x)$, we'll use the quotient rule:

$f'(x) = \frac{(3\cos x)'(x + 1)^3 - (3 \cos x)((x + 1)^3)'}{((x + 1)^3)^2}$

Calculating the derivatives:

$(3 \cos x)' = -3 \sin x$ and $((x + 1)^3)' = 3(x + 1)^2$

Substituting into the formula:

$f'(x) = \frac{-3 \sin x (x + 1)^3 - 3 \cos x (3(x + 1)^2)}{(x + 1)^6}$

Simplifying this we can factor out:

$f'(x) = \frac{g(x)}{(x + 1)^3}$

where $g(x) = -3 \sin x (x + 1)^3 - 9 \cos x (x + 1)^2$.