# What is the derivative of x + cos(x+y)=0?

##### 2 Answers
Jul 16, 2018

$y ' = \frac{1 - \sin \left(x + y\right)}{\sin} \left(x + y\right)$

#### Explanation:

Assuming you mean $y = y \left(x\right)$
we get by the chain rule

$1 - \sin \left(x + y\right) - y ' \sin \left(x + y\right) = 0$
so we get

$y ' = \frac{1 - \sin \left(x + y\right)}{\sin} \left(x + y\right)$ if $\sin \left(x + y\right) \ne 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}} - 1$

#### Explanation:

Given equation:

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

$\cos \left(x + y\right) = - x$

$x + y = \setminus \pi - \setminus {\cos}^{- 1} \left(x\right)$

$y = \setminus \pi - \setminus {\cos}^{- 1} \left(x\right) - x$

Differentiating above equation w.r.t. $x$ as follows

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{d}{\mathrm{dx}} \left(\setminus \pi - \setminus {\cos}^{- 1} \left(x\right) - x\right)$

$= 0 - \left(- \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}}\right) - 1$

$= \setminus \frac{1}{\setminus \sqrt{1 - {x}^{2}}} - 1$