Question

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

Answer 1

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

Explanation:

`-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}`