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Vi Nguyen · Apr 25, 2009, 09:37 PM

Can anyone tell me how to find the derivative of:

xy = tan(xy) and of

x² =arctan(y)/(1 + y²)

Very much appreciated, as soon as possible please, I really have no clue when there is both x and y inside the brackets!!

galactus · Apr 26, 2009, 05:22 AM

Use implicit differentiation and the chain rule.

$xy=tan(xy)$

$xy'+y=(xy'+y)sec^{2}(xy)$

$xy'+y=xy'sec^{2}(xy)+ysec^{2}(xy)$

$xy'-xy'sec^{2}(xy)=ysec^{2}(xy)-y$

$xy'(1-sec^{2}(xy))=y(sec^{2}(xy)-1)$

$xy'(1-sec^{2}(xy))=-y(1-sec^{2}(xy))$

$y'=\frac{-y}{x}$