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Question 78 of 99

Q.If f(x,y)=1x2+y2f(x,y) = \dfrac{1}{\sqrt{x^2+y^2}} then, prove that x∂f∂x+y∂f∂y=−fx\dfrac{\partial f}{\partial x} + y\dfrac{\partial f}{\partial y} = -f.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2019Subjective· 3mImportance★★★★★
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Computing the two partial derivatives directly and combining them as xfx+yfyx f_x+yf_y collapses back to −f-f, matching Euler's theorem for a degree −1-1 homogeneous function.

  1. Write f(x,y)=1x2+y2=(x2+y2)−1/2f(x,y)=\dfrac{1}{\sqrt{x^2+y^2}}=(x^2+y^2)^{-1/2}.
  2. Differentiate partially w.r.t. xx (treating yy constant): ∂f∂x=−12(x2+y2)−3/2⋅2x=−x(x2+y2)−3/2\dfrac{\partial f}{\partial x}=-\dfrac12(x^2+y^2)^{-3/2}\cdot 2x = -x(x^2+y^2)^{-3/2}.
  3. Similarly, ∂f∂y=−y(x2+y2)−3/2\dfrac{\partial f}{\partial y}=-y(x^2+y^2)^{-3/2}. …

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