If you only need a close approximation, you can approximate it linearly pretty easily.
Let f(x)=x^(1/2), then f'(x)=1/2 x^(-1/2).
Since .9 is pretty close to 1, and we can find f'(1)=1/2, we know that the slope of the line passing through (1,1) and tangent to the curve of f(x) is 1/2. Then, use y=mx+b to find the equation of the tangent line. we know m=1/2, and (1,1) is a point on the line. Plug this in, and find b=1/2. So, y=1/2x+1/2. Let x=.9, and we get y=.9/2+1/2, which is .95. That's the linear approximation.
Does anyone know how to input formulas in these forums? I'm new.