Let fxygu where ux2y2 and gu is a differentiable function in

Let f(x,y)=g(u) where u=x^2+y^2 and g(u) is a differentiable function in one variable. (?f/?x)^2+(?f/?y)^2-4u(?g/?u)^2=0

f(x,y) = g(x^2 + y^2)
f/x = g\'(x^2 + y^2) * 2x
(f/x)^2 = [g\'(x^2 + y^2)]^2 * 4x^2..........(1)
f/y = g\'(x^2 + y^2) * 2y
(f/y)^2 = [g\'(x^2 + y^2)]^2 * 4y^2..........(2)
g/u = g\'(x^2 + y^2)
(g/u)^2 = [g\'(x^2 + y^2)]^2
4u(g/u)^2 = 4(x^2 + y^2)[g\'(x^2 + y^2)]^2........(3)
apply operation (1) + (2) - (3)
(f/x)^2+(f/y)^2-4u(g/u)^2 = [g\'(x^2 + y^2)]^2 * 4x^2 + [g\'(x^2 + y^2)]^2 * 4y^2 + 4(x^2 + y^2)[g\'(x^2 + y^2)]^2
(f/x)^2+(f/y)^2-4u(g/u)^2 = 0
hence proved

$Let f(x,y)=g(u) where u=x^2+y^2 and g(u) is a differentiable function in one variable. (?f/?x)^2+(?f/?y)^2-4u(?g/?u)^2=0Solutionf(x,y) = g(x^2 + y^2) f/x = g\'($

Let fxygu where ux2y2 and gu is a differentiable function in

Solution

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