Linear approximation

Question

Use linear approximation or differentials to determine the value of sqrt(100.5)

Answer 1

Take the derivative of the square root function y = √ x, y^' = 1/(2√x), use x =100 as its the closest perfect square to 100.5. You should get y^' = 1/20, then use the slope-point form with the point (100, 10) and slop 1/20. The linear approx is L(x) = (1/20)(x-100) +10, plug in 100.5 to get 10.025.