# How do you differentiate sinx-siny=x-y-5?

Oct 29, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x - 1}{\cos y - 1}$

#### Explanation:

$\sin x - \sin y = x - y - 5$

Differentiate wrt $x$
$\therefore \frac{d}{\mathrm{dx}} \sin x - \frac{d}{\mathrm{dx}} \sin y = \frac{d}{\mathrm{dx}} x - \frac{d}{\mathrm{dx}} y - \frac{d}{\mathrm{dx}} 5$
$\therefore \cos x - \frac{d}{\mathrm{dx}} \sin y = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$

We can't differentiate $\sin y$ wrt $x$ but we can differentiate wet y using the chain rule (this is the implicit differentiation)

$\therefore \cos x - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \sin y = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \cos x - \frac{\mathrm{dy}}{\mathrm{dx}} \cos y = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$

We can now collect terms and factorise:
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \cos y - \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x - 1$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos y - 1\right) = \cos x - 1$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x - 1}{\cos y - 1}$