# How do you find dy/dx by implicit differentiation given y=sqrt(xy+1)?

Feb 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x y + 1}}{2 \sqrt{x y + 1} - x}$

#### Explanation:

Square both sides to eliminate the need to use the chain rule.

${y}^{2} = {\left(\sqrt{x y + 1}\right)}^{2}$

${y}^{2} = x y + 1$

Now use implicit differentiation and the product rule to find the derivative.

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x y + 1}}{2 \sqrt{x y + 1} - x}$

Hopefully this helps!