Question

implicit differentiation help.
Somebody here has got to know how to do this problem... >.< nobody has been able to get it thus far. :|

http://gyazo.com/53dfb49c3ca8d1a3b949b4f9f2afa67c

Answer 1

wow, talk about issuing a challenge :)

Answer 2

._.

Answer 3

where have you gotten so for

Answer 4

i've solved for dy/dx. plugged the x and y coordinates in and got -1.732. so that should be my slope for the tangent line. i plug in values to y-y1=m(x-x1). which would be y-1=-1.732(x+3sqrt3)

Answer 5

but when i set that equal to y, i dont get the right answer for some reason. doing something wrong

Answer 6

goku!! :)

Answer 7

hi :)

Answer 8

checking it now

Answer 9

\[\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}y'=0\]is a start

Answer 10

I got \(\large y' = \frac{\sqrt{3}}{3}\)

Answer 11

ye. got that far.

Answer 12

\(\large y-1 = \frac{\sqrt{3}}{3}(x + 3\sqrt{3})\)

Answer 13

just to compare.. i got y = -x/sqrt(3) -2 but he says its wrong .. i would like to see a different answer

Answer 14

cruffo y' shouldnt be negative ?

Answer 15

your right curffo. wonder what i was doing wrong. :l

Answer 16

So that probably solves it :)

Answer 17

darn negatives...

Answer 18

>.<

Answer 19

(2/3) * x^-(1/3) + (2/3) * y^-(1/3) * (dy/dx) = 0

dy/dx = -x^(-1/3) / y^(-1/3)

Answer 20

see pic

Answer 21

yes got it now .. when i plugged into x i plugged without the sign

Answer 22

wow thanks a lot. should have hlped me when you saw my problem the first time. :P

Answer 23

would have saved me an hr or so.

Answer 24

ha! what the fun in that.

Answer 25

LOL