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Mathematics 7 Online
OpenStudy (anonymous):

Implicit differentiation

OpenStudy (anonymous):

If you guys take the derivative with respect to x for the function \[2y^{3}-3xy+2=0\]

OpenStudy (anonymous):

do you guys get \[\frac{3y}{6y^{2}-3x}\]

hartnn (hartnn):

chain rule and product rul, if you know, this is a piece of cake :)

OpenStudy (anonymous):

When you evaluate it at y = 2 (NOT x=2) do you get \[\frac{2}{8-x}\]

hartnn (hartnn):

when y= 2, what does x equal ?

OpenStudy (anonymous):

nothing it just says evaluate at y = 2

hartnn (hartnn):

use this : 2y3−3xy+2=0 to find out

ganeshie8 (ganeshie8):

in ur derivative square terms should disappear

OpenStudy (anonymous):

in the denominator? why? It was 2y^{3} take the derivative and it becomes 6y^{2}?

hartnn (hartnn):

2*8-6x+2=0

hartnn (hartnn):

i think your derivative part is correct

ganeshie8 (ganeshie8):

i thought it was 2x^2... i need to go check doc :|

OpenStudy (anonymous):

wait I know what x is!! x = 3

hartnn (hartnn):

yes :)

OpenStudy (anonymous):

Its alright ganeshie so do I ..so do I

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