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heard of implicit differentiation ?
no i have never heard of it
you really just use the chain rule. Most of the time the derivative with respect to x is 1, so you don't think about the derivative of 3x^2 being 3x*1, but when you do implicit differentiation you take the derivative of 3y^2 to be 3y*(dy/dx)
u want to diff w..r.t x, right ?
6* not 3x or 3y by the way! lol
ok, so u know product rule, and chain rule, can u differentiate xy ?
no i can not lol i have never had xy together only y=x or something
but u know product rule right ? take x and y as different functions and diff. xy using product rule
like x'y+xy' and x' = ?
would it be y+x?
umm.....x'=1 right ? so diff. of xy will be y+xy' ok
but isnt y' also 1? so it would be y+x?
y' would have been 1 if you would have diff. w.r.t y as in d/dy of y =1 but here you are differentiating w.r.t x do its diff. would be dy/dx that is y' makes sense ?
ohhh okay! thank you, that makes more sense
and what will be diff. of y^2 then ?
since d/dx of x^2 = 2x d/dy of y^2 = 2y dy/dx = 2yy' ok?
okay so the diff of y^3 would be 3y²y' ?
x² + xy + y² = 1 soo 2x+(x+y')+2y'=0 is that right
2nd term ? y+xy'
oops yeah that is what i meant 2x+(y+xy')+2y'=0
yes, now find y' from there ?
how do i do that? lol
2x+(y+xy')+2yy' =0 -2x=y+xy' +2yy' -2x-y=xy'+2yy' -2x-y/2y=xy'/2y +y' uhm im stuck
2x+(y+xy')+2yy' =0 xy'+2yy' = -2x-y (x+2y)y' = -(2x+y) now ?
(x+2y)y' = -(2x+y) y'=-(2x+y)/(x+2y)
wasn't it easy ?
thank you! is that the entire thing finished?
yup., u got your y'
easier than i thought, thank you, i was just curious how to do a question like this, it isn't needed :P you are awesome!