# How do you differentiate y cos x = 1 + sin (xy)?

Aug 17, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left[\cos \left(x y\right) + \sin x\right]}{\cos x - x \cos \left(x y\right)}$

#### Explanation:

Always think of your chain, product and quotient rules when differentiating

$y \cos x = 1 + \sin \left(x y\right)$
Diff wrt $x$:
$\frac{\mathrm{dy}}{\mathrm{dx}} \cos x - y \sin x = \cos \left(x y\right) \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right]$ (apply your rules)
$\frac{\mathrm{dy}}{\mathrm{dx}} \cos x - y \sin x = y \cos \left(x y\right) + x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$ (simplify)
$\frac{\mathrm{dy}}{\mathrm{dx}} \cos x - x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y \cos \left(x y\right) + y \sin x$ (collect $\frac{\mathrm{dy}}{\mathrm{dx}}$)
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[\cos x - x \cos \left(x y\right)\right] = y \left[\cos \left(x y\right) + \sin x\right]$ (factorise)
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left[\cos \left(x y\right) + \sin x\right]}{\cos x - x \cos \left(x y\right)}$ VOILA