The curve C has equation $$y = \frac{(x^2 + 4)(x - 3)}{2x}, \quad x \neq 0$$ (a) Find $$\frac{dy}{dx}$$ in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

To find $\frac{dy}{dx}$, we will use the quotient rule. The function is given as:

$y = \frac{(x^2 + 4)(x - 3)}{2x}$

Let:

• $u = (x^2 + 4)(x - 3)$
• $v = 2x$

The quotient rule states that:

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

First, we need to calculate $\frac{du}{dx}$ and $\frac{dv}{dx}$:

1. Calculate $\frac{du}{dx}$:

   • Using the product rule on $u$:

   $\frac{du}{dx} = (x^2 + 4) \cdot 1 + (2x)(x - 3)$

   • Simplifying gives: $\frac{du}{dx} = x^2 + 4 + 2x^2 - 6x = 3x^2 - 6x + 4$

2. Calculate $\frac{dv}{dx}$:

   • $\frac{dv}{dx} = 2$

Now, substitute these into the quotient rule:

$\frac{dy}{dx} = \frac{2x(3x^2 - 6x + 4) - (x^2 + 4)(x-3)(2)}{(2x)^2}$

After simplifying the numerator and denominator, we find:

$\frac{dy}{dx} = \frac{12x^3 - 24x^2 + 16x - 2x^3 - 4x + 12}{4x^2}$

Combine and simplify further to arrive at:

$\frac{dy}{dx} = \frac{10x^3 - 24x^2 + 12}{4x^2} = \frac{5x^3 - 12x^2 + 6}{2x^2}$

Thus, the final answer in its simplest form is:

$\frac{dy}{dx} = \frac{5x^2 - 12x + 6}{2x}$