What is the slope of the tangent line of x/y^3-xy= C , where C is an arbitrary constant, at (1,2)?

1 Answer
Jul 27, 2018

Slope of the tangent line at (1,2) is -1.58

Explanation:

x/y^3-x y = C; (1,2) or

x*y^(-3)-x y = C; , differentiating we get,

y^(-3)+ x (-3 y^-4)y^'- y-x y^' = 0; or

y^'(-(3 x)/y^4 - x)= y- 1/y^3

:. y^'= (y- 1/y^3)/(-((3 x)/y^4 + x) ; Putting x=1 ,y=2

we get ,y^'= (2- 1/2^3)/(-((3*1)/2^4 + 1)) or y^'~~ -1.58

Slope of the tangent line at (1,2) is m ~~ -1.58 [Ans]