Question

let (f(x)=\frac{sqrt{x}-4}{sqrt{x}+4}). then,
(f(x)=\frac{4}{sqrt{x}(sqrt{x}+4)^2})
(f(2)=)

Explanation:

Step1: Substitute x = 2 into f'(x)

We know \(f'(x)=\frac{4}{\sqrt{x}(\sqrt{x} + 4)^2}\), substitute \(x = 2\).

Step2: Calculate the value

When \(x = 2\), \(\sqrt{x}=\sqrt{2}\). Then \(f'(2)=\frac{4}{\sqrt{2}(\sqrt{2}+ 4)^2}\).
First, expand \((\sqrt{2}+4)^2=(\sqrt{2})^2 + 2\times4\times\sqrt{2}+4^2=2 + 8\sqrt{2}+16=18 + 8\sqrt{2}\).
So \(f'(2)=\frac{4}{\sqrt{2}(18 + 8\sqrt{2})}=\frac{4}{18\sqrt{2}+16}\).
Rationalize the denominator: Multiply numerator and denominator by \(18\sqrt{2}-16\).
The numerator is \(4\times(18\sqrt{2}-16)=72\sqrt{2}-64\).
The denominator is \((18\sqrt{2}+16)(18\sqrt{2}-16)=(18\sqrt{2})^2-16^2=648 - 256 = 392\).
So \(f'(2)=\frac{72\sqrt{2}-64}{392}=\frac{18\sqrt{2}-16}{98}=\frac{9\sqrt{2}-8}{49}\).

Answer:

\(\frac{9\sqrt{2}-8}{49}\)