4. (a) Differentiate with respect to x (i) $x^{2}e^{x^{2}}$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 5

To differentiate (y = \frac{\cos(2x)}{3x}), we again apply the quotient rule. If (u = \cos(2x)) and (v = 3x), then:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Calculating the derivatives:

• (\frac{du}{dx} = -2\sin(2x)) using the chain rule.
• (\frac{dv}{dx} = 3).

Substituting these values:

$\frac{dy}{dx} = \frac{3x(-2\sin(2x)) - \cos(2x)(3)}{(3x)^2} = \frac{-6x\sin(2x) - 3\cos(2x)}{9x^2}$

To find (\frac{dy}{dx}), we will use implicit differentiation. Starting from (x = 4 \sin(2y + 6)), we differentiate both sides with respect to x:

1. The left-hand side: (\frac{dx}{dx} = 1).
2. The right-hand side using the chain rule:
   • Let (u = 2y + 6), then (\frac{du}{dx} = 2\frac{dy}{dx}), so: $\frac{dx}{dx} = 4 \cos(2y + 6) \cdot 2 \frac{dy}{dx} \implies 1 = 8 \cos(2y + 6) \frac{dy}{dx}$

Solving for (\frac{dy}{dx}) gives:

$\frac{dy}{dx} = \frac{1}{8 \cos(2y + 6)}$

Next, from the original equation for (x), we can express (\cos(2y + 6)) in terms of x:

Using (4 \sin(2y + 6) = x), we apply the Pythagorean identity to obtain:

$\cos(2y + 6) = \sqrt{1 - \sin^2(2y + 6)} = \sqrt{1 - \left( \frac{x}{4} \right)^2} \implies \frac{dy}{dx} = \frac{1}{2 \sqrt{16 - x^2}}$.