# What is the slope of the tangent line of  (xy^2-3xy)/(sqrt(yx)-xy) =C , where C is an arbitrary constant, at (2,5)?

$- 0.653281301$

#### Explanation:

Given implicit function $\setminus \frac{x {y}^{2} - 3 x y}{\setminus \sqrt{y x} - x y} = C$

Differentiating the above function wrt $x$ on both the sides we get

$\setminus \frac{\left(\setminus \sqrt{y x} - x y\right) \setminus \frac{d}{\mathrm{dx}} \left(x {y}^{2} - 3 x y\right) - \left(x {y}^{2} - 3 x y\right) \setminus \frac{d}{\mathrm{dx}} \left(\setminus \sqrt{y x} - x y\right)}{{\left(\setminus \sqrt{y x} - x y\right)}^{2}} = 0$

$\left(\setminus \sqrt{y x} - x y\right) \left(2 x y y ' + {y}^{2} - 3 x y ' - 3 y\right) - \left(x {y}^{2} - 3 x y\right) \left(\frac{1}{2} \setminus \sqrt{\frac{x}{y}} y ' + \frac{1}{2} \setminus \sqrt{\frac{y}{x}}\right) = 0 \setminus \quad \left(\setminus \forall \setminus x y \setminus \ne 0 , 1\right)$

$\left(2 \left(\setminus \sqrt{x y} - x y\right) \left(2 x y - 3 x\right) - \setminus \sqrt{\frac{x}{y}} \left({x}^{2} y - 3 x y\right)\right) y ' = \setminus \sqrt{\frac{y}{x}} \left({x}^{2} y - 3 x y\right) - 2 \left(\setminus \sqrt{x y} - x y\right) \left({y}^{2} - 3 y\right)$

$y ' = \setminus \frac{\setminus \sqrt{\frac{y}{x}} \left({x}^{2} y - 3 x y\right) - 2 \left(\setminus \sqrt{x y} - x y\right) \left({y}^{2} - 3 y\right)}{2 \left(\setminus \sqrt{x y} - x y\right) \left(2 x y - 3 x\right) - \setminus \sqrt{\frac{x}{y}} \left({x}^{2} y - 3 x y\right)}$

Now, setting $x = 2 , y = 5$ in above equation, we get the slope $y '$ of the tangent line

$y ' = \setminus \frac{\setminus \sqrt{\frac{5}{2}} \left({2}^{2} \setminus \cdot 5 - 3 \left(2\right) \left(5\right)\right) - 2 \left(\setminus \sqrt{2 \setminus \cdot 5} - 2 \setminus \cdot 5\right) \left({5}^{2} - 3 \setminus \cdot 5\right)}{2 \left(\setminus \sqrt{2 \setminus \cdot 5} - 2 \setminus \cdot 5\right) \left(2 \setminus \cdot 2 \setminus \cdot 5 - 3 \setminus \cdot 2\right) - \setminus \sqrt{\frac{2}{5}} \left({2}^{2} \setminus \cdot 5 - 3 \setminus \cdot 2 \setminus \cdot 5\right)}$

$y ' = \frac{120.9430585}{-} 185.1316702$

$y ' = - 0.653281301$