Edexcel Pure Paper 1 2024 Question 5

Start with the function:

f(x) = \frac{2x – 3}{x^2 + 4}

Part (a)

This is quotient rule. No way around it.

Write it down first — don’t try to do it from memory mid-step:

\left(\frac{u}{v}\right)' = \frac{v u' – u v'}{v^2}

Let:

• u = 2x – 3 → derivative is 2
  
  
• v = x^2 + 4 → derivative is 2x
  
  

Now plug straight in:

f'(x) = \frac{(x^2 + 4)(2) – (2x – 3)(2x)}{(x^2 + 4)^2}

At this point, slow down a bit. Most mistakes happen here, not earlier.

Expand each bit separately.

First part:

2(x^2 + 4) = 2x^2 + 8

Second part:

(2x – 3)(2x) = 4x^2 – 6x

Now subtract. This is where signs go wrong if you rush:

2x^2 + 8 – (4x^2 – 6x)

Careful:

= 2x^2 + 8 – 4x^2 + 6x

= -2x^2 + 6x + 8

So:

f'(x) = \frac{-2x^2 + 6x + 8}{(x^2 + 4)^2}

That matches the required form.

So:

• a = -2
  
  
• b = 6
  
  

c = 8