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

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Answer 1 · 2015-12-27T07:47:43

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

#-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))#

#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 }#