Differentiate with respect to $x$, giving your answer in its simplest form, (a) $x^2 \ln(3x)$ (b) $\frac{\sin 4x}{x^3}$ - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 6

To differentiate the function $g(x) = \frac{\sin 4x}{x^3}$, we will use the quotient rule:

1. Identify the numerator and denominator:

   • Let $u = \sin(4x)$ and $v = x^3$.

2. Differentiate each part:

   • $u' = \frac{d}{dx}(\sin(4x)) = 4\cos(4x)$
   • $v' = \frac{d}{dx}(x^3) = 3x^2$

3. Apply the quotient rule: $g'(x) = \frac{u'v - uv'}{v^2}$ $g'(x) = \frac{4\cos(4x) \cdot x^3 - \sin(4x) \cdot 3x^2}{(x^3)^2}$

4. Simplify the expression: $g'(x) = \frac{4x^3 \cos(4x) - 3x^2 \sin(4x)}{x^6}$ $g'(x) = \frac{4\cos(4x) - 3 \frac{\sin(4x)}{x}}{x^3}$

Thus, the final solution for (b) in its simplest form is: $g'(x) = \frac{4\cos(4x) - 3 \frac{\sin(4x)}{x}}{x^3}$