Question

How do you implicitly differentiate `-1=xytan(x/y)`?

Answer 1

`frac{dy}{dx} = frac{y}{x}*frac{ 2xcsc(frac{2x}{y}) + y }{ xcot(x/y) - y }`

Explanation:

`-1 = xytan(x/y)`

Differentiate both sides w.r.t. `x`.

`frac{d}{dx}(-1) = frac{d}{dx}(xytan(x/y))`

`0 = frac{d}{dx}(xytan(x/y))`

Use the Product Rule, Chain Rule, and Quotient Rule.

`frac{d}{dx}(xytan(x/y)) = xyfrac{d}{dx}(tan(x/y)) + tan(x/y)frac{d}{dx}(xy)`

`= xyfrac{d}{dx}(tan(x/y)) + tan(x/y)(yfrac{d}{dx}(x) + xfrac{d}{dx}(y))`

`= xysec^2(x/y)frac{d}{dx}(x/y) + tan(x/y)(y + xfrac{dy}{dx})`

`= xysec^2(x/y)frac{yfrac{d}{dx}(x) - xfrac{d}{dx}(y)}{y^2} + tan(x/y)(y +xfrac{dy}{dx})`

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

`= xsec^2(x/y) - frac{x^2}{y}frac{dy}{dx} + ytan(x/y)+xtan(x/y)frac{dy}{dx}`

`= xsec^2(x/y) + ytan(x/y) + (xtan(x/y) - frac{x^2}{y})frac{dy}{dx} = 0`

`frac{dy}{dx} = - frac{ xsec^2(x/y) + ytan(x/y) }{ xtan(x/y) - frac{x^2}{y} }`

`= frac{y}{x}*frac{ 2xcsc(frac{2x}{y}) + y }{ xcot(x/y) - y }`