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For the equation f(x,y) = arctan (x/y)

how do you calculate it’s first and second order partial derivative?

 

i know the way to find it but for this question I keep getting it wrong, so i wish someone can help me with it. Thanks. 

 Apr 27, 2019

Best Answer 

 #1
avatar+5226 
+1

\(f(x,y) = \arctan\left(\dfrac x y\right)\\ f_x = \dfrac{1}{1+\left(\dfrac x y\right)^2}\cdot \dfrac 1 y = \dfrac{y}{x^2+y^2}\)

 

\(f_y = \dfrac{1}{1+\left(\dfrac x y \right)^2}\cdot \dfrac{-x}{y^2} = -\dfrac{x}{x^2+y^2}\)

 

\(f_{xx} = -\dfrac{y}{(x^2+y^2)^2}\cdot 2x = -\dfrac{2xy}{(x^2+y^2)^2}\)

 

\(f_{yy} = \dfrac{2xy}{(x^2+y^2)^2}\)

 

\(f_{xy}=f_{yx} =\dfrac{ 1(x^2+y^2)-2y(y)}{(x^2+y^2)^2} = \dfrac{x^2-y^2}{(x^2+y^2)^2}\)

.
 Apr 27, 2019
 #1
avatar+5226 
+1
Best Answer

\(f(x,y) = \arctan\left(\dfrac x y\right)\\ f_x = \dfrac{1}{1+\left(\dfrac x y\right)^2}\cdot \dfrac 1 y = \dfrac{y}{x^2+y^2}\)

 

\(f_y = \dfrac{1}{1+\left(\dfrac x y \right)^2}\cdot \dfrac{-x}{y^2} = -\dfrac{x}{x^2+y^2}\)

 

\(f_{xx} = -\dfrac{y}{(x^2+y^2)^2}\cdot 2x = -\dfrac{2xy}{(x^2+y^2)^2}\)

 

\(f_{yy} = \dfrac{2xy}{(x^2+y^2)^2}\)

 

\(f_{xy}=f_{yx} =\dfrac{ 1(x^2+y^2)-2y(y)}{(x^2+y^2)^2} = \dfrac{x^2-y^2}{(x^2+y^2)^2}\)

Rom Apr 27, 2019
 #2
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Thank you so much. 

Guest Apr 28, 2019

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