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

Jul 27, 2018

Slope of the tangent line at $\left(1 , 2\right)$ 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

${y}^{'} \left(- \frac{3 x}{y} ^ 4 - x\right) = y - \frac{1}{y} ^ 3$

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

we get ,${y}^{'} = \frac{2 - \frac{1}{2} ^ 3}{- \left(\frac{3 \cdot 1}{2} ^ 4 + 1\right)} \mathmr{and} {y}^{'} \approx - 1.58$

Slope of the tangent line at $\left(1 , 2\right)$ is $m \approx - 1.58$ [Ans]