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Question Number 129039 by oustmuchiya@gmail.com last updated on 12/Jan/21

p a r t i a l l y D i f f e r e n t i a t e f u n c t i o n

f (x, y) = 2 x^{- 2} y + {x y}^{3} + \frac{2}{x^{2} y}

Answered by liberty last updated on 12/Jan/21

f (x, y) = {2 x}^{- 2} y + {xy}^{3} + {2 x}^{- 2} y^{- 1}

{\begin{cases} \frac{\partial f}{\partial x} = - {4 x}^{- 3} y + y^{3} - {4 x}^{- 3} y^{- 1} \\ \frac{\partial f}{\partial y} = {2 x}^{- 2} + {3 xy}^{2} - {2 x}^{- 2} y^{- 2} \end{cases}

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