# How do you differentiate ycosx^2-y^2=xy?

Jan 7, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(2 x \sin \left({x}^{2} + 1\right)\right)}{\cos \left({x}^{2}\right) - x - 2 y}$.

#### Explanation:

Firstly, let's apply the derivative operator to both sides, I'll try my best to colour code the use of the product rule:

$\frac{d}{\mathrm{dx}} \left[\textcolor{red}{y \cos \left({x}^{2}\right)} \textcolor{t e a l}{- {y}^{2}}\right] = \frac{d}{\mathrm{dx}} \left[\textcolor{p u r p \le}{x} \textcolor{\mathmr{and} a n \ge}{y}\right]$

$\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left({x}^{2}\right) + y \cdot \left(- \sin \left({x}^{2}\right) \cdot 2 x\right)} \textcolor{t e a l}{- 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}} = \textcolor{p u r p \le}{1} \cdot \textcolor{\mathmr{and} a n \ge}{y} + \textcolor{p u r p \le}{x} \cdot \textcolor{\mathmr{and} a n \ge}{\frac{\mathrm{dy}}{\mathrm{dx}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left({x}^{2}\right) - 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x y \cdot \sin \left({x}^{2}\right) + y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos \left({x}^{2}\right) - 2 y - x\right) = 2 x y \cdot \sin \left({x}^{2}\right) + y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y \cdot \sin \left({x}^{2}\right) + y}{\cos \left({x}^{2}\right) - 2 y - x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(2 x \sin \left({x}^{2} + 1\right)\right)}{\cos \left({x}^{2}\right) - x - 2 y}$.