Someone argues: "both partial derivatives of f(x,y)= xy^2 x^2+y^2 , (x,y) (0,0)

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Calculus of change and accumulation · Calculus of change and accumulation

Conditions of Validity, Ch Differentiability, the implicit function theorem and the licence to change variables

Problem

Someone argues: "both partial derivatives of f(x,y)=xy2x2+y2,(x,y)≠(0,0),f(0,0)=0,f(x,y)=\frac{xy^2}{x^2+y^2},\qquad (x,y)\ne(0,0),\qquad f(0,0)=0, vanish at the origin, and ff is continuous there, so ff must be differentiable at the origin." Test this claim. ((a)) Show fx(0,0)=fy(0,0)=0f_x(0,0)=f_y(0,0)=0, and show ff is continuous at the origin. ((b)) Write R(h,k)=f(h,k)−[fx(0,0)h+fy(0,0)k]R(h,k) = f(h,k) - [f_x(0,0)h+f_y(0,0)k], evaluate R(h,k)/h2+k2R(h,k)/\sqrt{h^2+k^2} in plane polar coordinates h=rcos⁡θh=r\cos\theta, k=rsin⁡θk=r\sin\theta, and use the result to decide whether ff is differentiable at the origin.

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