The functions g and h are such that g(x) = \sqrt{2x - 5} h(x) = \frac{1}{x} (a) Find g(16) (b) Find hg'(x) Give your answer in terms of x in its simplest form - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 2

First, we need to determine h'(x):

Since (h(x) = \frac{1}{x}), using the power rule:

$h'(x) = -\frac{1}{x^2}$.

Next, compute hg'(x):

$hg'(x) = g(h(x)) \cdot h'(x)$
Substituting for g and h:

$= g\left(\frac{1}{x}\right) \cdot -\frac{1}{x^2}$

Now substituting (\frac{1}{x}) into g:

$g\left(\frac{1}{x}\right) = \sqrt{2\left(\frac{1}{x}\right) - 5} = \sqrt{\frac{2}{x} - 5}$.

Taking the product:

$hg'(x) = \sqrt{\frac{2}{x} - 5} \cdot -\frac{1}{x^2}$

Thus, the answer is:

$hg'(x) = -\frac{\sqrt{\frac{2}{x} - 5}}{x^2}$.