Please help with the derivative problem: f(x)= 2/√x, a= 1/9. What is f'(a)

f'(a)=limit of (f(a+h)-f(a))/h as h approaches 0. f(x)= 2/√x, a= 1/9. What is f'(a)? The algebra involved in this problem is quite complicated, please helps

f'(a) = (f(a+h) - f(a))/h

for f(x) = 2/√x, a= 1/9:

f(a) = 2/√(1/9)
f(a) = 2/(1/3)
f(a) = 6

f(a+h) = 2/√((1/9) +h)
f(a+h) = 2/√((1+9h)/9)
f(a+h) = 2/(√(1+9h)/3)
f(a+h) = 6/√(1+9h)

f'(a) = (f(a+h) - f(a))/h
f'(a) = ((6/√(1+9h)) - 6)/h
f'(a) = (6/√(1+9h)) - 6√(1+9h)/√(1+9h))/h
f'(a) = (6(1-√(1+9h))/√(1+9h))/h

Now, as x -> 0 , √(1+x) -> 1 + x/2 (try this with a calculator and numbers such as 1.01, 1.006, etc.),
so as h -> 0, √(1+9h) -> 1+9h/2

f'(a) = (6(1-(1+9h/2))/(1+9h/2))/h
f'(a) = ((-6 9h/2)/(1+9h/2))/h
f'(a) = -27h/(h(1+9h/2)
f'(a) = -27/(1+9h/2)

As h -> 0, f'(a) -> -27