# What is the slope of the tangent line of  (xy-1/(yx))(xy+1/(xy)) =C , where C is an arbitrary constant, at (1,5)?

Jan 15, 2016

I got $- 5$

#### Explanation:

We are told that the point $\left(1 , 5\right)$ lies on $\left(x y - \frac{1}{y x}\right) \left(x y + \frac{1}{x y}\right) = C$.

So we can find $C$ to be $25 - \frac{1}{25}$. (We won't actually need this.)

As written, this looks tedious to differentiate implicitly, so let's rewrite it as:

${x}^{2} {y}^{2} - \frac{1}{{x}^{2} {y}^{2}} = C$.

Or better, still,

${x}^{4} {y}^{4} - 1 = C {x}^{2} {y}^{2}$. So

$4 {x}^{3} {y}^{4} + 4 {x}^{4} {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 C x {y}^{2} + 2 C {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

$2 {x}^{2} {y}^{3} + 2 {x}^{3} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = C y + C x \frac{\mathrm{dy}}{\mathrm{dx}}$

Solving for the derivative gets us

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 {x}^{2} {y}^{3} - C y}{2 {x}^{3} {y}^{2} - C x}$

Evaluating at $\left(1 , 5\right)$, we get

dy/dx]_"(1,5)" = -(2(5)^3-5C)/(2(5)^2-C) = -(5(2(5)^2-C))/(2(5)^2-C)=-5

Bonus

Regardless of the constant, at $x = 1$, the slope of the tangent will be $- y$