Derivatives and Tangents

Question

I'm totally confused.  The problem is: use the fact that the derivative of the function g(x)=square root is g`(x) equals 1over 2square rooto f x to find lthe equation of the tangent line to the graph of g(x) at the point x=81.  What is the correct answer and why is my answer of 1/41x+8/2 is not right?

Answer 1

y' = 1/[2√x]

y'(81) = m, slope of tangent to curve at point where x=81.

m = 1/[2•9] = 1/18

Point of tangency is (81,9) and equation of the tangent, using Point-slope form of line, is 

y - 9 = (1/18)(x - 81).

Proceed from there (if you wish to simplify).