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 5 - 2014 - Paper 6

1. Start by differentiating (y = (x^2 + x) \ln 2x): Use the product rule: $\frac{dy}{dx} = (x^2 + x) \frac{d}{dx}(\ln 2x) + \ln 2x \frac{d}{dx}(x^2 + x)$

2. Find (\frac{d}{dx}(\ln 2x) = \frac{1}{2x} * 2 = \frac{1}{x}): This gives: $\frac{dy}{dx} = (x^2 + x) \cdot \frac{1}{x} + \ln 2x (2x + 1)$

3. Substitute (x = \frac{e}{2}) into the derivative: $\frac{dy}{dx} = \left(\left(\frac{e}{2}\right)^2 + \frac{e}{2}\right) \cdot \frac{1}{\frac{e}{2}} + \ln(2 * \frac{e}{2}) \left(2 * \frac{e}{2} + 1\right)$

4. Calculate each term to find the specific value.

To show (f'(x) = \frac{g(x)}{( x + 1 )^3}):

1. Differentiate (f(x)) using the quotient rule: $f'(x) = \frac{(x + 1)^3(\frac{d}{dx}(3cos\,x)) - 3cos\,x \frac{d}{dx}((x + 1)^3)}{(x + 1)^6}$

2. Calculate the derivatives:

   • (\frac{d}{dx}(3cos,x) = -3sin,x)
   • (\frac{d}{dx}((x + 1)^3) = 3(x + 1)^2)

3. Substitute these into the expression: $f'(x) = \frac{(x + 1)^3(-3sin\,x) - 3cos\,x \cdot 3(x + 1)^2}{(x + 1)^6}$

4. Simplify the numerator to identify (g(x)): $g(x) = -3(x + 1)(sin\,x) - 9cos\,x$

5. Finally, express (f'(x) = \frac{g(x)}{( x + 1 )^3}$$.