7. (a) Show that $h(x) = \frac{2x}{x^{2} + 5} - \frac{4}{x^{2} + 5} - \frac{18}{(x+5)(x+2)}.$ (b) Hence, or otherwise, find $h'(x)$ in its simplest form - Edexcel - A-Level Maths Pure - Question 21 - 2013 - Paper 1

To show that the given expression for $h(x)$ is correct, we start by finding a common denominator for the three fractions.

The common denominator is $(x^{2} + 5)(x + 5)(x + 2)$. We can rewrite each fraction:

1. The first term: $\frac{2x}{x^{2} + 5} = \frac{2x(x + 5)(x + 2)}{(x^{2} + 5)(x + 5)(x + 2)}$

2. The second term: $\frac{4}{x^{2} + 5} = \frac{4(x + 5)(x + 2)}{(x^{2} + 5)(x + 5)(x + 2)}$

3. The third term: $\frac{18}{(x + 5)(x + 2)} = \frac{18(x^{2} + 5)}{(x + 5)(x + 2)(x^{2} + 5)}$

Now combining these fractions, we have:

$h(x) = \frac{2x(x + 5)(x + 2) - 4(x + 5)(x + 2) - 18(x^{2} + 5)}{(x^{2} + 5)(x + 5)(x + 2)}$

Expanding and simplifying the numerator results in:

$2x(x^2 + 7x + 10) - (4x^2 + 28x + 40) - (18x^2 + 90)$

After simplifying, we observe that:

$2x - \frac{(x^{2} + 5)(x + 5)(x + 2)}{(x^{2} + 5)(x + 5)(x + 2)}$

This shows that the proposed expression for $h(x)$ is indeed valid.