2007-07-09 03:20:35 · 4 answers · asked by Demetria S 1 in Science & Mathematics ➔ Mathematics
z'(w) = -1/2(3w -2)^(-3/2)
2007-07-09 03:25:27 · answer #1 · answered by Runa 7 · 0⤊ 0⤋
you just need to diffrentiate it with respect to the variable "w".
since z= 1/ (3w-2)^(1/2)
therefore
dz/dw = d/dw (3w-1)^(-1/2)
= -1/2 * (3w-1)^(-1/2-1) * 3
= -3/2* (3w-1)^(-3/2) [ answer]
2007-07-09 10:37:26 · answer #2 · answered by vicky 7 2 · 1⤊ 0⤋
Let u =3w-2
then dz/dw = dz/du*du/dw
Chain rule
Now du/dw = 3
and z =1/u^1/2 so dz/du = -1/2*u^-3/2
Subsituting back
Dz/dw = -3/2*(3w-2)^(-3/2)
2007-07-09 10:29:47 · answer #3 · answered by nyphdinmd 7 · 1⤊ 0⤋
z = 1/sqrt(3w-2) = sqrt(3w-2)^(-1) = (3w-2)^(-1/2)
now just use the power rule, and the chain rule:
(-1/2)*(3w-2)^(-3/2)*(3), which, if you don't like negative and/or fractional exponents, is:
-3 / [2*(sqrt(3w-2)^3)]
2007-07-09 10:28:07 · answer #4 · answered by grompfet 5 · 1⤊ 0⤋