a) To show that \( f'(x) = \frac{10 – 2x^2}{(x^2 + 5)^2} \):
Let \( f(x) = \frac{2x}{x^2 + 5} \).
Use the quotient rule: If \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{v \times u’ – u \times v’}{v^2} \).
Set \( u = 2x \), \( v = x^2 + 5 \).

Derivatives:
\( u’ = 2 \)
\( v’ = 2x \)

Apply quotient rule:
\( f'(x) = \frac{(x^2 + 5) \times 2 – 2x \times 2x}{(x^2 + 5)^2} \)
\( = \frac{2(x^2 + 5) – 4x^2}{(x^2 + 5)^2} \)
\( = \frac{2x^2 + 10 – 4x^2}{(x^2 + 5)^2} \)
\( = \frac{10 – 2x^2}{(x^2 + 5)^2} \)

Thus:
\( f'(x) = \frac{10 – 2x^2}{(x^2 + 5)^2} \) [4]

b) To find \( \int \frac{2x}{x^2 + 5} \, \text{d}x \):
Use substitution: Let \( u = x^2 + 5 \).
Then \( \text{d}u = 2x \, \text{d}x \), so \( \frac{2x \, \text{d}x}{x^2 + 5} = \frac{\text{d}u}{u} \).

Integrate:
\( \int \frac{1}{u} \, \text{d}u = \ln u + c \)
Substitute back:
\( \ln (x^2 + 5) + c \)

Thus:
\( \int \frac{2x}{x^2 + 5} \, \text{d}x = \ln (x^2 + 5) + c \) [4]

c) To find the value of \( q \):
The area of the shaded region is given by:
\( \int_{\sqrt{5}}^{q} \frac{2x}{x^2 + 5} \, \text{d}x = \ln 7 \)

Use the antiderivative from part (b):
\( \int \frac{2x}{x^2 + 5} \, \text{d}x = \ln (x^2 + 5) \)
Evaluate the definite integral:
\( \left[ \ln (x^2 + 5) \right]_{\sqrt{5}}^{q} = \ln (q^2 + 5) – \ln ((\sqrt{5})^2 + 5) \)
\( = \ln (q^2 + 5) – \ln (5 + 5) \)
\( = \ln (q^2 + 5) – \ln 10 \)
\( = \ln \left( \frac{q^2 + 5}{10} \right) \)

Set equal to the given area:
\( \ln \left( \frac{q^2 + 5}{10} \right) = \ln 7 \)
\( \frac{q^2 + 5}{10} = 7 \)
\( q^2 + 5 = 70 \)
\( q^2 = 65 \)
\( q = \sqrt{65} \)

Note: Since the shaded region is to the right of \( x = \sqrt{5} \), \( q > \sqrt{5} \), so \( q = \sqrt{65} \) (positive root).
Award A0 for \( q = \pm \sqrt{65} \).

Thus:
\( q = \sqrt{65} \) [7]