2007-01-15 05:32:11 · 6 answers · asked by Kevin E 1 in Science & Mathematics ➔ Mathematics
If you didn't make a mistake and the statement is correct, then you mean the constant function f(x) = sqrt(2) for every real x. Then f'(x) = 0.
2007-01-15 05:55:08 · answer #1 · answered by Steiner 7 · 0⤊ 0⤋
The derivative of any constant is 0.
Are you trying to use Newton's Method to approximate it? In this case you should make a function with a root at Sqrt(2) and then use the derivative of that function--not the derivative of Sqrt(2) itself. For instance, y=x^2-2 should work nicely.
In this case, y'=2x, so Newton's Method says that a guess X will evolve into the next guess, X-(X^2-2)/(2X) = (X^2+2)/(2X).
An initial guess of 1 becomes 3/2, then 17/12, then 577/408, and so on.
2007-01-15 06:05:32 · answer #2 · answered by Doc B 6 · 0⤊ 0⤋
The derivative of a constant is 0. The square root of 2 is a constant. Therefore its derivative is zero.
2007-01-15 05:37:08 · answer #3 · answered by 1ofSelby's 6 · 1⤊ 0⤋
constants (numbers without a variable) have a derivative of 0.
2007-01-15 05:41:25 · answer #4 · answered by just me 2 · 0⤊ 0⤋
That would be 0 because there is no x after the number.
2007-01-15 05:40:12 · answer #5 · answered by Anonymous · 0⤊ 0⤋
it is 0 as it is a constant
2007-01-15 05:35:28 · answer #6 · answered by raj 7 · 1⤊ 0⤋