Find y' of y= 3x^-4 - 1/x^2
2007-10-12 04:39:23 · 6 answers · asked by jewel7962002 1 in Science & Mathematics ➔ Mathematics
y' = -12x^-5 + 1/x^3
2007-10-12 04:45:18 · answer #1 · answered by Anonymous · 0⤊ 0⤋
if y = x^n then y' = n x^(n-1)
y = 3x^(-4) - x^(-2)
y' = -12x^(-5)+ 2x^(-3)
2007-10-12 04:49:08 · answer #2 · answered by norman 7 · 0⤊ 0⤋
y= 3x^-4 - 1/x^2
y' = dy/dx = 3*(-4)*x^(-4 - 1) - (-2)x^(-2 -1)
or y' = -12x^-5 +2x^-3
2007-10-12 04:47:09 · answer #3 · answered by ib 4 · 0⤊ 0⤋
y' = 12x^-5+2/x^3
2007-10-12 04:48:58 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
dy/dx = 3(-4)x^(-3) - (-2)x^(-3)
because 1/x^2 = x^(-2)
-12/x^3+2/x^3
=-10/x^3
2007-10-12 04:51:14 · answer #5 · answered by cidyah 7 · 0⤊ 0⤋
y'=12/x^-5 + 2/x^3
but you really should be doing your homework yourself
2007-10-12 04:47:45 · answer #6 · answered by thatchickwithredhair 3 · 0⤊ 0⤋