4. (a) Differentiate with respect to x (i) $x^2 e^{x^2 + 2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 5

To differentiate $g(x) = \frac{\cos(2x)}{3x}$, we apply the quotient rule:

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

Where $u = \cos(2x)$ and $v = 3x$. Calculating:

• $u' = -2 \sin(2x)$ (using chain rule)
• $v' = 3$

Now substituting into the quotient rule:

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

Simplifying gives us:

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

From the equation $x = 4 \sin(2y + 6)$, we differentiate both sides with respect to x:

$1 = 4 \cos(2y + 6) \cdot 2\frac{dy}{dx}$

Thus:

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

We need to express this in terms of $x$. From $x = 4 \sin(2y + 6)$, we rearrange:

$\sin(2y + 6) = \frac{x}{4}$

Using the Pythagorean identity:

$\cos^2(2y + 6) = 1 - \sin^2(2y + 6) = 1 - \left(\frac{x}{4}\right)^2$

Thus:

$\cos(2y + 6) = \sqrt{1 - \frac{x^2}{16}}$

Therefore, substituting into our derivative gives:

$\frac{dy}{dx} = \frac{1}{8 \sqrt{1 - \frac{x^2}{16}}}$