If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$ 
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$ 

My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x: 
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore
$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}}  $$
Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$ 
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.
 A: The answer is indeed $\frac13$. To check this, let's go another way: set $g=x-\frac2x$, as you did. Then express $x$ through $g$: 
$$x=\frac12\left(g\pm\sqrt{g^2+8}\right).$$
Then
$$f(g)=\sqrt{\frac12\left(g\pm\sqrt{g^2+8}\right)-1}.$$
Taking its derivative, we can find that it's real at $g=1$ for $\pm\to+$, and we get
$$f'(1)=\frac13.$$
(The other solution with $\pm\to-$ is $-\frac{i}{6\sqrt2}$).
Why does your source say the answer is $1$? Maybe it's just a mistake of evaluating $f(1)$ instead of $f'(1)$, since $f(1)=1$.
A: Differentiate both sides to get $f'(x-\frac{2}{x})(1+\frac{2}{x^2})=\frac{1}{2\sqrt{x-1}}$.  Then use $x=2$ to evaluate,  $f'(1)=\frac{1}{3}$.
A: Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Also $f(g(x))=f'(g(x)).g'(x)$ by the chain rule.
for $ g(x)=x-\frac {2}{x}$
$\Rightarrow g'(x) = 1+\frac {2}{x^2}$
for $f'(x)=\frac{1}{2} \frac {1}{\sqrt{(x-1)}}$
$f(g(x))=\left(1+\frac {2}{x^2}\right) \left(\frac {1}{2\sqrt{(x-1)}}\right)$
$\Rightarrow \frac {-5}{8}$ $\text{by differentiating and putting x=2 for already evaluated g(x) for x=1}\\$
A: Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Taking the positive value of $x$
Therefore $f(1) = \sqrt {x-1}$ and 
Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$
$$\Rightarrow \frac{1}{2}\cdot(x-1)^{\frac{-1}{2}}$$
Putting the positive value $2$, we get 
$f'(1) = \frac{1}{2}$
