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

Apr 8, 2016

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

#### Explanation:

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

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

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

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

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