what is the derivative of (1/x)?
2007-09-17 10:16:35 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1/x is just x^(-1), so the same "power rule" applies here. Moving the -1 down to the front and subtracting 1 from the exponent gives -x^(-2).
2007-09-17 10:20:30 · answer #1 · answered by Anonymous · 0⤊ 1⤋
f (x) = x^(-1)
f `(x) = (-1) x ^( - 2)
f `(x) = (-1) / x ²
2007-09-20 06:24:03 · answer #2 · answered by Como 7 · 1⤊ 0⤋
-1 / x^2
how ? 1/x = x^(-1)
the derivative is (-1)x^(-2) = -1/x^2
2007-09-17 10:23:33 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋
It's
f (x) = x^(-1)
Then you derivate it as a normal exponent
f `(x) = (-1) x ^( - 2)
Same as
f `(x) = (-1) / x ²
2015-02-26 01:01:14 · answer #4 · answered by Gianluca Lardo 1 · 0⤊ 0⤋
Use the power rule:
1/x = x^(-1)
-1x^(-2) = -1/(x^2)
2007-09-17 10:23:14 · answer #5 · answered by sayamiam 6 · 0⤊ 0⤋
By chain rule d/dx f(g(x)) = f'(g(x))g'(x), we have f(x) = g(x)^(½) and g(x) = 1 - x. Taking the derivative of those functions gives f'(x) = (½)g(x)^(-½) and g'(x) = -x. Put that altogether, we obtain: dy/dx = (0.5)(1 - x)^(0.5 - 1)(-1) => (-½)(1 - x)^(-½) => -1/√(1 - x) [In terms of radicals] I hope this helps!
2016-05-17 07:03:43 · answer #6 · answered by Anonymous · 0⤊ 0⤋
-(1/(x^2))
Think of 1/x to be the same as x^(-1).
The dervative of x^(-1) is -x^(-2).
Using the format of the original number, this would be -(1/(x^2))
2007-09-17 10:23:10 · answer #7 · answered by Larry C 3 · 0⤊ 0⤋
1/x = x^-1
-1x^(-2)
2007-09-17 10:21:28 · answer #8 · answered by ccw 4 · 0⤊ 1⤋
-1/x²
Doug
2007-09-17 10:29:35 · answer #9 · answered by doug_donaghue 7 · 0⤊ 0⤋