# How do you implicitly differentiate -1=sinxy ?

Feb 29, 2016

$y ' = - \frac{y}{x}$

#### Explanation:

$- 1 = \sin \left(x y\right)$

Differentiate implicitly with respect to $x$:

$D \left(- 1\right) = D \left(\sin x y\right)$

Use the chain rule on $d \frac{\sin \left(x y\right)}{\mathrm{dx}} \Rightarrow$

$0 = \cos \left(x y\right) \times \left(x y ' + y\right)$

$\therefore 0 = \cos \left(x y\right) \times x y ' + \cos \left(x y\right) \times y$

$\therefore \cancel{\cos \left(x y\right)} \times x y ' = \cancel{- \cos \left(x y\right)} \times y$

$\therefore x y ' = - y$

$\therefore y ' = - \frac{y}{x}$