# What is the implicit derivative of xy+ysqrt(xy-x) =8 ?

Aug 16, 2016

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

#### Explanation:

first lets get the derivatives with respect to $x$ and $y$
$\frac{d}{\mathrm{dx}} = y + \frac{y \left(y - 1\right)}{2 \sqrt{x y - x}}$

$\frac{d}{\mathrm{dx}} = \frac{y 2 \sqrt{x y - x} + y \left(y - 1\right)}{2 \sqrt{x y - x}}$

$\frac{d}{\mathrm{dy}} = x + \sqrt{x y - x} + x y \left(\frac{1}{2 \sqrt{x y - x}}\right)$

$\frac{d}{\mathrm{dy}} = \frac{2 x \sqrt{x y - x}}{2 \sqrt{x y - x}} + \frac{2 \left(x y - x\right)}{2 \sqrt{x y - x}} + \frac{x y}{2 \sqrt{x y - x}}$

$\frac{d}{\mathrm{dy}} = \frac{2 x \sqrt{x y - x} + 2 x \left(y - 1\right) + \left(x y\right)}{2 \sqrt{x y - x}}$

next note that
d/dx=d/dy⋅dy/dx or equivalently d/dy⋅dy/dx -d/dx=0

$\frac{2 x \sqrt{x y - x} + 2 x \left(y - 1\right) + \left(x y\right)}{2 \sqrt{x y - x}} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{y 2 \sqrt{x y - x} + y \left(y - 1\right)}{2 \sqrt{x y - x}} = 0$
$\frac{2 x \sqrt{x y - x} + 2 x \left(y - 1\right) + \left(x y\right)}{2 \sqrt{x y - x}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y 2 \sqrt{x y - x} + y \left(y - 1\right)}{2 \sqrt{x y - x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y 2 \sqrt{x y - x} + y \left(y - 1\right)}{2 \sqrt{x y - x}} \cdot \frac{2 \sqrt{x y - x}}{2 x \sqrt{x y - x} + 2 x \left(y - 1\right) + \left(x y\right)}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y 2 \sqrt{x y - x} + y \left(y - 1\right)}{2 x \sqrt{x y - x} + 2 x \left(y - 1\right) + \left(x y\right)}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y 2 \sqrt{x y - x} + {y}^{2} - y}{2 x \sqrt{x y - x} + 3 x y - 2 x}$