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Mathematics 11 Online
OpenStudy (photon336):

question

OpenStudy (photon336):

@marcelie

OpenStudy (photon336):

yeah so basically an equation f(x,y) = 0 defines y implicitly as a function of x. the domain of that function consist of those x for which there is a unique y such that f(x,y) = 0

OpenStudy (marcelie):

oh okay thats when you use dy/dx right ?

OpenStudy (photon336):

yeah i'm still trying to figure this out. i think it's when you take the derivative and the solutions are such that x,y = 0

OpenStudy (marcelie):

oh okay is that when you have two equations = to each other and do dy/dx for y's ?

OpenStudy (photon336):

so let's look at the equation of a circle. and c is a constant. \[x^{2}+y^{2}= c \]

OpenStudy (photon336):

first thing we do is take the derivative of both sides

OpenStudy (photon336):

\[\frac{ dy }{ dx }*x^{2}+\frac{ dy }{ dx }y = \frac{ dy }{ dx}*c\] the derivative of a constant is zero. \[\frac{ dy }{ dx }*x^{2}+\frac{ dy }{ dx }y = 0\]

OpenStudy (marcelie):

hmm would it look something like this ? |dw:1462683301381:dw|

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