The answer is (-2)/(x^3). I'm trying to do this w/ the diff. form., but after writing it out and using common demons to get a fraction on the numerator of the fraction, I have no idea what to do. Thanks.
2007-05-22 11:04:03 · 3 answers · asked by The Ghetto David Hume 3 in Science & Mathematics ➔ Mathematics
thx but could someone do this by way of diff formula, thats how I'm going to need how to do it on the test.
2007-05-22 11:20:48 · update #1
ok thx gta
2007-05-22 11:26:35 · update #2
If you mean using difference quotient, here it is:
http://img483.imageshack.us/my.php?image=diffquotientjv6.png
2007-05-22 11:20:32 · answer #1 · answered by Anonymous · 1⤊ 0⤋
If f(x) = x^n
f ` (x) = n.x^(n - 1)
f (x) = x^(-2)
f ` (x) = (-2).x^(- 3)
f ` (x) = (-2) / x³
2007-05-23 06:17:40 · answer #2 · answered by Como 7 · 0⤊ 0⤋
use the quotient rule:
dy/dx [f(x)]/[g(x)] = [g(x)(f(x))' - f(x)(g(x))']/[(g(x))^2]
dy/dx = [(x^2)(1)' - (1)(x^2)']/[(x^2)^2]
dy/dx = [(2x)(0)-(1)(2x)]/(x^4)
dy/dx = -2x/(x^4)
dy/dx = -2/(x^3)
Oh sorry, I didn't realize you had to do it a certain way.
2007-05-22 18:09:48 · answer #3 · answered by Anonymous · 0⤊ 0⤋