9. (a) Show that $f''(x) = \frac{(3 - x^3)^2}{x^2}$, where $x \neq 0$ where $A$ and $B$ are constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 1

The first derivative of $f(x)$ can be found by using the power rule:

Starting from:

$f''(x) = 9x^2 + A + Bx^2$

When we integrate:

$f'(x) = \int(9x^2 + A + Bx^2) \ dx = 3x^3 + Ax + \frac{B}{3}x^3 + C$

where $C$ is the integration constant.

Given that the point $(-3, 10)$ lies on the curve, we need to use the derivatives to find the function:

Integrating $f'(x)$ gives:

$f(x) = \int(3x^3 + Ax + \frac{B}{3}x^3 + C) \ dx$

This leads to:

$f(x) = \frac{3}{4}x^4 + \frac{A}{2}x^2 + \frac{B}{12}x^4 + Cx + D$

Using the point $(-3, 10)$, substituting $x = -3$ into the equation will help solve for constants $C$ and $D$.