What is the slope of the tangent line of #xy^2-(1-x/y)^2= C #, where C is an arbitrary constant, at #(1,-1)#?

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Answer 1 · 2016-11-04T18:50:02

slope=#dy/dx=-1/2#

To find the slope of the tangent line you have to find the derivative first then plug in the point for x and y

#xy^2-(1-x/y)^2=C#

#y^2+2xydy/dx-2(1-x/y)*(-1/y)-2(1-x/y)(x/y^2) dy/dx=0#

#y^2+2xydy/dx+(2y-2x)/y^2 +(2x^2-2xy)/y^3 dy/dx =0#

#2xydy/dx+(2x^2-2xy)/y^3 dy/dx =-y^2-(2y-2x)/y^2#

#dy/dx(2xy+(2x^2-2xy)/y^3)=-y^2-(2y-2x)/y^2#

#dy/dx((2xy^4+2x^2-2xy)/y^3)=(-y^4-2y+2x)/y^2#

#dy/dx=(-y^4-2y+2x)/y^2 *y^3/(2xy^4+2x^2-2xy)#

#dy/dx=(-y^5-2y^2+2xy)/(2xy^4+2x^2-2xy)#

#dy/dx=(-(-1)^5-2(-1)^2+2(1)(-1))/(2(1)(-1)^4+2(1)^2-2(1)(-1)#

#dy/dx=-3/6=-1/2#