Given y = \sqrt{x} + \frac{4}{\sqrt{x}} + 4, \quad x > 0 find the value of \frac{dy}{dx} when x = 8, writing your answer in the form a\sqrt{2}, where a is a rational number. - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 1

Now, we substitute x = 8 into the derivative:

$\frac{dy}{dx} = \frac{1}{2\sqrt{8}} - \frac{2}{8^{\frac{3}{2}}}$

Calculating each term:

• First term: \sqrt{8} = 2\sqrt{2} \Rightarrow 2\sqrt{8} = 4\sqrt{2}\Rightarrow \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}}.
• Second term: \quad 8^{\frac{3}{2}} = 8\sqrt{8} = 8\cdot 2\sqrt{2} = 16\sqrt{2} \Rightarrow \frac{2}{8^{\frac{3}{2}}} = \frac{2}{16\sqrt{2}} = \frac{1}{8\sqrt{2}}.

Putting it all together: $\frac{dy}{dx} = \frac{1}{4\sqrt{2}} - \frac{1}{8\sqrt{2}} = \left(\frac{2}{8\sqrt{2}} - \frac{1}{8\sqrt{2}}\right) = \frac{1}{8\sqrt{2}}$

The answer is: $\frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}$

Thus, where a is a rational number, a = \frac{1}{16}.