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Derivatives

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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

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

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

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

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

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

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Apr 27, 2019

#1
+5768
+1

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

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

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

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

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

Rom Apr 27, 2019
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Thank you so much.

Guest Apr 28, 2019