If #f(x,y)=x^2tan^(-1)(y/x)-y^2tan^(-1)(x/y),x ne 0,y ne 0# and if #f(x,y)=0, x=0=y#, prove that #(partial^2f)/(partialx partialy)=(x^2-y^2)/(x^2+y^2)#?

1 Answer
Steve M
Feb 17, 2018

We have:

# f(x,y) = { (0, x=y=0), (x^2tan^(-1)(y/x)-y^2tan^(-1)(x/y), "otherwise") :}#

And so we can compute the first partial derivatives by application of the chain and product rules for #x,y ne 0# as follows:

# f_x = (partial f)/(partial x) #
# \ \ \ = {x^2 * 1/(1+(y/x)^2) * (-y/x^2) + 2x \ tan^(-1)(y/x)} #
# \ \ \ \ - {y^2 * 1/(1 +(x/y)^2) * 1/y + 0}#

# \ \ \ = -y/(1+(y/x)^2) + 2x \ tan^(-1)(y/x) - y/(1 +(x/y)^2) #

# \ \ \ = 2x \ tan^(-1)(y/x) -(x^2y)/(x^2+y^2) - y^3/(y^2 +x^2) #

# \ \ \ = 2x \ tan^(-1)(y/x) - (y(x^2+y^2))/(y^2 +x^2) #

# \ \ \ = 2x \ tan^(-1)(y/x) - y #

Similarly, we get:

# f_y = (partial f)/(partial y) #
# \ \ \ = {x^2 * 1/(1+(y/x)^2) 1/x + 0} #
# \ \ \ \ - {y^2 * 1/(1 +(x/y)^2) * (-x/y^2) + 2ytan^(-1)(x/y)}#

# \ \ \ = x/(1+(y/x)^2) + x/(1 +(x/y)^2) - 2y \ tan^(-1)(x/y)#

# \ \ \ = x^3/(x^2+y^2) + (xy^2)/(y^2 +x^2) - 2y \ tan^(-1)(x/y)#

# \ \ \ = (x(x^2+y^2))/(x^2+y^2) - 2y \ tan^(-1)(x/y)#

# \ \ \ = x- 2y \ tan^(-1)(x/y) #

So, we have:

# f_x = 2x \ tan^(-1)(y/x) - y \ # and # \ f_y = x- 2y \ tan^(-1)(x/y)#

So then, The Second Partial Cross-Derivative we seek is:

# f_(xy) = (partial^2 f)/(partial x partial y) = (partial)/(partial x) (partial f)/(partial y)#

# \ \ \ = (partial)/(partial x) (x- 2y \ tan^(-1)(x/y))#

# \ \ \ = 1 - 2{y * 1/(1+(x/y)^2) * 1/y + 0}#

# \ \ \ = 1 - 2/(1+(x/y)^2) #

# \ \ \ = 1 - 2y^2/(y^2+x^2) #

# \ \ \ = ((y^2+x^2) - 2y^2)/(y^2+x^2) #

# \ \ \ = (x^2 - y^2)/(x^2+y^2) \ \ \ # QED

Note:

If we were to compute #f_(yx)# we would find that

# f_(yx) = (x^2 - y^2)/(x^2+y^2) #

The second partial cross derivatives are identical due to the continuity of #f(x,y)#.

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