# How do you differentiate -2y=xcosy-xy?

Aug 18, 2016

dy/dx=-(y-cosy)^2/(2(cosy+ysiny).

#### Explanation:

We rewrite the given eqn. as $- 2 y = x \left(\cos y - y\right) , \mathmr{and} , x = \frac{2 y}{y - \cos y}$.

Diff.ing w.r.t. $y$ both sides of this eqn., we have,

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{d}{\mathrm{dy}} \left\{\frac{2 y}{y - \cos y}\right\} = 2 \frac{d}{\mathrm{dy}} \left\{\frac{y}{y - \cos y}\right\}$.

Using Quotient Rule for Diffn. on the R.H.S., we get,

$\frac{\mathrm{dx}}{\mathrm{dy}} = 2 \left[\frac{\left(y - \cos y\right) \cdot \frac{d}{\mathrm{dy}} \left(y\right) - y \cdot \frac{d}{\mathrm{dy}} \left(y - \cos y\right)}{y - \cos y} ^ 2\right]$

$= 2 \left[\frac{\left(y - \cos y\right) \left(1\right) - \left(y\right) \left(1 + \sin y\right)}{y - \cos y} ^ 2\right]$

$= 2 \left\{\frac{y - \cos y - y - y \sin y}{y - \cos y} ^ 2\right\}$

$= - 2 \left\{\frac{\cos y + y \sin y}{y - \cos y} ^ 2\right\}$.

Therefore, dy/dx=1/(dx/dy)=-(y-cosy)^2/(2(cosy+ysiny).

Enjoy Maths.!