There's a different way to get to this, without using calculus. If we call x = sqrt(67), then the first approximation is n = sqrt(64) = 8. We then use the form

There's a different way to get to this, without using calculus. If we call x = sqrt(67), then the first approximation is n = sqrt(64) = 8. We then use the formula for the difference of squares x^2 - n^2 = (x-n)(x+n) to get x-n = (x^2-n^2)/(x+n), or x = n + (x^2-n^2)/(n+x). In our case x = 8 + 3/(8+x). This is circular, but we can just plug in the right hand side the first approximation for x, which is 8, and we end up with 8 + 3/16. The beauty of this formula is that you can continue to iterate. Once you have x ~ 8.1875, you can plug again in the right hand side and get x ~ 8 + 3/16.1875, which is exact to the first 4 decimals. Of course, you can continue, and you end up with the (aptly named) continued fraction method for the square root [1]. [1] https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion https://en.wikipedia.org/wiki/Methods_of_computing_square_ro...