totally confuse
xy^3 + sqr root(6xy)=12
I am basically confuse with the (square root of 6xy--)
any help would be appreciate
Thanks
2007-08-22 02:38:09 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
i start with
3xy^2 dy/dx +y^3 +6^1/2*x^1/2*y^1/2=0
6^1/2 would that give me sqr root of (3)
and should i multiple that by x^1/2*y^1/2 which is the product rule ...
2007-08-22 02:58:00 · update #1
Supposing you want to determine y', it's as follows:
Since the right had side is a constant, it's derivative is 0. So, acording to the rules for differentiating products and square rotots, we get
x (3y^2) y' + y^3 + sqrt(6) (x y' + y)/(2 sqrt(xy) = 0 So,
y' (6 x y^2 sqrt(xy)) + 2 y^3 sqrt(xy) + (2sqrt(6) x) y' + sqrt(6) y = 0
y' = - ( 2 y^3 sqrt(xy) + sqrt(6) y)/( 6 x y^2 sqrt(xy) + 2sqrt(6) x)
Double check, I may have a made a mistake. This exercise is not interesting at all, just algebra.
2007-08-22 02:57:00 · answer #1 · answered by Steiner 7 · 0⤊ 0⤋
Maths stands for Mathematical Anti Telharsic Harfatum Septomin.
2007-08-22 02:47:20 · answer #2 · answered by Morty 5 · 0⤊ 1⤋
treat it as:
xy^3 + (6xy)^(1/2) = 12
2007-08-22 02:42:54 · answer #3 · answered by miggitymaggz 5 · 0⤊ 1⤋
make sqrt(6xy) = (6xy)^(1/2)
2007-08-22 02:50:17 · answer #4 · answered by naijagunner 4 · 0⤊ 1⤋