Question

x^3+y^3+3axy=0
dy/dx=?

Answer 1

\[\LARGE 3x^{2}+3y^{2} \frac{dy}{dx}-3ax=0\]
this gives..finally..
\[\LARGE \frac{dy}{dx}=\frac{x(a-x)}{y^{2}} \]

I doubt this because im quite unsure about the derivative of 3axy,we are differentiating wrt x or y? if its x then 3ay(dx//dx) not sure..

Answer 2

@hartnn

Answer 3

you'll need product rule for 3axy

Answer 4

I thought product rule was used for function times function

Answer 5

thats a variable,isn't it

Answer 6

y is the function of x, x is the function of x.

Answer 7

:O

Answer 8

(xy)' = x'y+xy' = y+xy'

Answer 9

so how will we find for 3axy?

Answer 10

3a (y+xy')

Answer 11

\[\LARGE 3ay(\frac{dx}{dx}) |||| 3ax(\frac {dy}{dx}) \]

Answer 12

dx/dx=1

Answer 13

yeah!

Answer 14

and i assume ||| is +

Answer 15

yes

Answer 16

thats it?:O

Answer 17

so, dy/dx will change..

Answer 18

what do u mean?

Answer 19

i mean recalculate dy/dx

Answer 20

\[\LARGE \frac{dy}{dx}= \frac{-3(ay+x^{2})}{3(y^{2}-ax)} \]

Answer 21

-ax ?

Answer 22

sorry +
printing mistake :P

Answer 23

then ok.

Answer 24

:D