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/sci/ - Science & Math


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9516836 No.9516836 [Reply] [Original]

I don't understand, aren't zeros of a derivative supposed to be local minimums/maximums of a curve?

>> No.9516854
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9516854

>>9516836
>y = sin(y - x)

>> No.9516864

this thread is LITERALLY why we have /sqt/

>> No.9516879

>>9516854
What's wrong with implicit differentiation?

>> No.9516971

>>9516879
d/dx(sin(y-x)) = (dy/dx - 1)cos(y - x)

Setting this equal to y creates a diffeq which I'm pretty sure desmos doesn't kniw how to solve. And anyway the zeros of it wouldn't be local extrema because (dy/dx - 1)cos(y - x) = 0 doesn't imply that dy/dx = 0. Nice bait tho

>> No.9516994

>>9516971
>anyway the zeros of it wouldn't be local extrema because (dy/dx - 1)cos(y - x) = 0 doesn't imply that dy/dx = 0

but that's exactly what it implies as deriving both sides of y = sin( y - x ) gives dy/dx = cos( y - x )*( dy/dx - 1 )

>> No.9517129

>>9516994
You have y = (dy/dx - 1)cos(y - x), not dy/dx = (dy/dx - 1)cos(y - x). The difference is crucial. But anyway I have a feeling that Desmos treated y as independent of x when taking the derivative, giving y = -sin(y-x).

>> No.9517136

Have you not taken multivariate calculus? Min/max for multivariable equations are defined differently.

1.) Take the partial derivative of f with respect to x
2.) Take the partial derivative of f with respect to y
Set each equal to zero and solve for x and y. This f(x,y) is the min, max, or a saddle point.

>> No.9517373

>>9517129
>You have y = (dy/dx - 1)cos(y - x), not dy/dx = (dy/dx - 1)cos(y - x). The difference is crucial.
dy/dx = cos( y - x ) * ( dy/dx - 1 ) is not particularly meaningful as that is exactly equivalent to y = sin( y - x )

>>9517136
thanks for the explanation anon, i'll delete the thread now

>> No.9517374

>>9517373
>Error: You cannot delete a post this old.
ffs