implicit differentiate should function exactly the same as always ; Just remember that where (d/dx)x=1, (d/dx)y=y' (derivative of y with respect to x, when y is a function of x)
so if you come across a y² (or any power), you need to use the chain rule :
(d/dx)y² = 2y(d/dx)y=2yy'
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Ill do a random example ; if you need help on a specific one then post it

y=1+x²y³
(d/dx)y=(d/dx)(x²y³)
y'=2xy³+x²(3y²y')
y'-3x²y²y'=2xy³
y'(1-3x²y²)=2xy³
y'=2xy³/'(1-3x²y²)
This post was edited by Taxidermy on Oct 26 2011 12:32am