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thinay · Aug 13, 2009, 09:56 AM

Sure! No problem with me.. :)

Unknown008 · Aug 13, 2009, 09:58 AM

Thank you! :) :D :cool:

galactus · Aug 13, 2009, 04:07 PM

4.) $y = tan(x+y)$
ans: $-csc^{2}(x+y) = -(1 + y^{-2})$
my ans: $sec^{2}(x+y)(1 + dy/dx)$

This one we can do with implicit differentiation.

$y=tan(x+y)$

Chain rule:

Derivative of $tan(u)=sec^{2}(u)$

Derivative, implicitly, of x+y is 1+y'.

$y'-(1+y')sec^{2}(x+y)=0$

$y'=\frac{sec^{2}(x+y)}{1-sec^{2}(x+y)}$

$y'=-csc^{2}(x+y)$

You just didn't simplify further and solve for y'.