Question

Derivative with 3 variables?

Answer 1

\[\large x^\frac{2}{3} +y^\frac{2}{3} =a^\frac{2}{3} \]

Answer 2

A generally represents a CONSTANT. I doubt that's a third variable in your problem. :D
Derivative of a constant issss?

Answer 3

00000000000000000000000000000000000000000

Answer 4

You missed a couple zeros, but yes that's correct :o

Answer 5

can we cut the powers directly?

Answer 6

lol

Answer 7

Cut them? :o

Answer 8

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

Answer 9

\[\sqrt{\frac{3}{2}}x+\sqrt{\frac{3}{2}}y \frac{dy}{dx}=0 \]

Answer 10

\[x+y \frac{dx}{dx}=0\]

Answer 11

\[\frac{dy}{dx}=\frac{-x}{y}\]

Answer 12

:O pro maths!!

Answer 13

:")

Answer 14

\[\large \frac{2}{3}x^{\frac{-1}{2}} +\frac{2}{3}y^{\frac{-1}{2}}\frac{dy}{dx}=0\]This step looked good. I'm not sure what happened after that :O

Answer 15

1/2 meant square root and minus meant reciprocal so did that :D

Answer 16

oh DAM SORRY

Answer 17

But the exponent is only being applied to x :O not the constant in front :O Silly DSL, such unreliable internet access, upgrade to cable D:

Answer 18

DSL and DLS hha :D

Answer 19

XD

Answer 20

\[\frac{2}{3\sqrt{x}}+\frac{2}{3\sqrt{y}}=0\]

Answer 21

MISSING DY/DX!!*

Answer 22

Yesss :O scary!

Answer 23

\[\frac{dy}{dx}=\frac{9 \sqrt{x} \sqrt{y}}{ \sqrt{x} +\sqrt{y}}\]

Answer 24

something like this perhaps O.o

Answer 25

Something like this perhaps? :o
\[\large \frac{2}{3\sqrt{x}}+\frac{2}{3\sqrt{y}}\frac{dy}{dx}=0 \qquad \rightarrow \qquad \frac{dy}{dx}=-\frac{\left(\frac{2}{3\sqrt x}\right)}{\left(\frac{2}{3\sqrt y}\right)}\]
There is some more simplification possible it looks like though :D

Answer 26

I got
\[\frac{dy}{dx}=\frac{3xy}{\sqrt{3}(y+x)}\]

Answer 27

Oh I just noticed, these are suppose to be cube roots, not square roots right?
2/3 - 1 = -1/3

Answer 28

oh yes :P

Answer 29

\[\large 2(\frac{1}{3 \sqrt[3]{x}}+\frac{1}{3 \sqrt[3]{y}} ) =0\]

Answer 30

oh and take the LCM

Answer 31

I think it might help to solve this if you leave them as fractional exponents maybe :O
\[\large \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0\]Multiplying both sides by 3/2 gives us,\[\large x^{-1/3}+y^{-1/3}\frac{dy}{dx}=0\]Subtracting the x term from both sides gives us,\[\large y^{-1/3}\frac{dy}{dx}=-x^{-1/3}\]

Answer 32

\[\frac{3 \sqrt[3]{x}+3 \sqrt[3]{y}}{3 \sqrt[3]{x}}dy/dx=0\]

Answer 33

yeah same thing^^

Answer 34

y/x ?

Answer 35

\[\large \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}\]Should get something like this I think. Hmm.

Answer 36

\[\large \frac{dy}{dx}=\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\]

Answer 37

oooooooo :D

Answer 38

Yay team.

Answer 39

holdon

Answer 40

http://www.wolframalpha.com/input/?i=x%5E%282%2F3%29+%2By%5E%282%2F3%29%3D0+find+dy%2Fdx

cheers :D