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)?

1 Answer
Nov 4, 2016

slope=dy/dx=-1/2

Explanation:

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