By differentiation we get:
$f'(x)=\frac{-(x-1)^{-3/2}}{2}$
We can then substitute $x=2$ to find the gradient of the funtion at $(2,1)$,
$f(2)=\frac{-(2-1)^{-3/2}}{2}=\frac{-1}{2}$
So the gradient is $\frac{-1}{2}$.
Now we know the slope and a point on the line, therefore we can calculate the equation of the line. The equation of a straight line is always in the form $y=mx +c$, using this fact and substituting in the gradient we just found we get:
$y=\frac{-1}{2}x+c$, luckily we also know a point on the line to substitute in to find $c$.
Using $(2,1)$:
$1=\frac{-1}{2}2+c$
rearranges to $c=2$ so, $y=\frac{-1}{2}x+2$
Multiplying everything by $2$ will give us: $2y=x+4$ or alternatively; $x+2y-4=0$