# How do you implicitly differentiate csc(x^2+y^2)=e^-x-y ?

$y ' = \frac{- 2 {x}^{2} \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) + {e}^{- x}}{2 y \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) - 1}$

#### Explanation:

$\csc \left({x}^{2} + {y}^{2}\right) = {e}^{- x} - y$

differentiate both sides of the equation with respect to x

$\frac{d}{\mathrm{dx}} \left(\csc \left({x}^{2} + {y}^{2}\right)\right) = \frac{d}{\mathrm{dx}} \left({e}^{- x} - y\right)$

$- \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({e}^{- x}\right) - \frac{d}{\mathrm{dx}} \left(y\right)$

$- \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) \left(2 {x}^{2} + 2 y \cdot y '\right) = \left({e}^{- x}\right) \cdot \left(- 1\right) - y '$

$- \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) \left(2 {x}^{2} + 2 y \cdot y '\right) = - {e}^{- x} - y '$

Multiplication then transposition

(2x^2(-csc (x^2+y^2)*cot (x^2+y^2))+e^(-x)=
$2 y \cdot y ' \cdot \left(\csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right)\right) - y '$

factor out the y'

$- 2 {x}^{2} \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) + {e}^{- x} =$
$\left[2 y \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) - 1\right] y '$

Divide both sides by $\left[2 y \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) - 1\right]$

$y ' = \frac{- 2 {x}^{2} \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) + {e}^{- x}}{2 y \cdot \csc \left({x}^{2} + {y}^{2}\right) \cdot \cot \left({x}^{2} + {y}^{2}\right) - 1}$

God bless....I hope the explanation is useful