Question

Please help me do this question genius minds

Answer 1

`\text{Given that,}\\\\\theta =\tan^(-1) \left(\frac y x \right)\\\\\\(\partial \theta)/(\partial x) = (\partial )/(\partial x) \tan^(-1) \left(\frac yx \right)\\ \\\\~~~~~~=\frac{1}{1 +\tfrac{y^2}{x^2}} \cdot y (\partial )/(\partial x) \left(\frac 1x \right)\\\\\\~~~~~~=(yx^2)/(x^2 +y^2) \cdot (-1)x^(-2)\\ \\\\~~~~~~=-(yx^2)/(x^2(x^2 +y^2))\\\\\\~~~~~~=- (y)/(x^2 +y^2)\\\\\\`

`(\partial^2\theta)/( \partial x^2) = -(\partial )/( \partial x) \left( (y)/(x^2+y^2) \right)\\\\\\~~~~~~=-y(\partial)/(\partial x) \left(\frac 1{x^2 +y^2} \right)\\\\\\~~~~~~=y (-1)(-1) (x^2 +y^2)^(-2) (\partial )/(\partial x)(x^2 +y^2)\\\\\\~~~~~~=(y)/((x^2 +y^2)^2) \cdot (2x+0)\\\\\\~~~~~~=(2xy)/((x^2 +y^2)^2)`

`\text{Similarly,}~~\\ \\(\partial \theta)/(\partial y )= (x)/(x^2 +y^2)\\\\\\(\partial^2 \theta)/(\partial y^2 ) = -(2xy)/((x^2 +y^2)^2)\\\\\\\text{Now,}\\\\(\partial^2 \theta)/(\partial x^2 )+(\partial^2 \theta)/(\partial y^2 )\\\\\\=(2xy)/((x^2 +y^2)^2) -(2xy)/((x^2 +y^2)^2)\\\\\\=0`