2019-02-28T10:57:15-05:00
Obtain all the first and second order partial derivatives of the function:
f( x, y) =ln( x^2 + y^2)
1
2019-03-01T08:44:51-0500
f ( x , y ) = l n ( x 2 + y 2 ) f( x, y) =ln( x^2 + y^2) f ( x , y ) = l n ( x 2 + y 2 )
∂ f ∂ x ( x , y ) = 2 x x 2 + y 2 \dfrac{\partial f} {\partial x}( x, y) =\dfrac{2x}{ x^2 + y^2} ∂ x ∂ f ( x , y ) = x 2 + y 2 2 x
∂ f ∂ y ( x , y ) = 2 y x 2 + y 2 \dfrac{\partial f} {\partial y}( x, y) =\dfrac{2y}{ x^2 + y^2} ∂ y ∂ f ( x , y ) = x 2 + y 2 2 y
∂ 2 f ∂ x 2 ( x , y ) = 2 y 2 − x 2 ( x 2 + y 2 ) 2 \dfrac{\partial^2 f} {\partial x^2}( x, y) =2\dfrac{y^2-x^2}{ (x^2 + y^2)^2} ∂ x 2 ∂ 2 f ( x , y ) = 2 ( x 2 + y 2 ) 2 y 2 − x 2
∂ 2 f ∂ y ∂ x ( x , y ) = − 4 x y ( x 2 + y 2 ) 2 \dfrac{\partial^2 f} {\partial y \partial x}( x, y) =\dfrac{-4xy}{ (x^2 + y^2)^2} ∂ y ∂ x ∂ 2 f ( x , y ) = ( x 2 + y 2 ) 2 − 4 x y
∂ 2 f ∂ x ∂ y ( x , y ) = − 4 x y ( x 2 + y 2 ) 2 \dfrac{\partial^2 f} {\partial x \partial y}( x, y) =\dfrac{-4xy}{ (x^2 + y^2)^2} ∂ x ∂ y ∂ 2 f ( x , y ) = ( x 2 + y 2 ) 2 − 4 x y
∂ 2 f ∂ y 2 ( x , y ) = 2 x 2 − y 2 ( x 2 + y 2 ) 2 \dfrac{\partial^2 f} {\partial y^2}( x, y) =2\dfrac{x^2-y^2}{ (x^2 + y^2)^2} ∂ y 2 ∂ 2 f ( x , y ) = 2 ( x 2 + y 2 ) 2 x 2 − y 2
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