# How do you find dy/dx for the equation y^2 - xy = - 5?

May 3, 2018

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

#### Explanation:

Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$. However, some functions $y$ are written implicitly as functions of $x$.

So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

One such equation is ${y}^{2} - x y = - 5$.

Here differentiating ${y}^{2}$ w.r.t. $y$ gives $2 y$ and hence differential of ${y}^{2}$ w.r.t. $x$ is $2 y \frac{\mathrm{dy}}{\mathrm{dx}}$. For $x y$ we treat it as a product of $x$ and $y$ and use product formula. Differential of $- 5$ is of course $0$ as it is constant. Hence differential of ${y}^{2} - x y = - 5$ is given by

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

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

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