How do you implicitly differentiate $-1=xy-tan(x-y)$ ?

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Answer 1 · 2016-02-07T02:29:05

$frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}$

$-1 = xy - tan(x-y)$

$frac{d}{dx}(-1) = frac{d}{dx}(xy - tan(x-y))$

$0 = frac{d}{dx}(xy) - frac{d}{dx}(tan(x-y))$

$= xfrac{d}{dx}(y) + yfrac{d}{dx}(x) - sec^2(x-y)frac{d}{dx}(x-y)$

$= xfrac{dy}{dx} + y - sec^2(x-y)(1-frac{dy}{dx})$

Now we just have to "shift terms" to make $frac{dy}{dx}$ the subject of the formula.

$sec^2(x-y) - y = (sec^2(x-y) + x)frac{dy}{dx}$

$frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}$

How do you implicitly differentiate -1=xy-tan(x-y) -1=xy-tan(x-y) ?