If # u = x^2 arctan(x/y) - y^2 arctan(y/x) # then find # (partial ^2u)/(partial x partial y) #?

1 Answer
Steve M
Oct 21, 2017

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

Explanation:

We have:

# u = x^2 arctan(x/y) - y^2 arctan(y/x) #

And we seek:

# (partial ^2u)/(partial x partial y) #

Let us first find the partial derivative wrt #y# by applying the chain rule and the product rule:

# (partial u)/(partial y) = x^2 ((partial)/(partial y)arctan(x/y)) - y^2 ((partial)/(partial y) arctan(y/x)) - ((partial)/(partial y)y^2) arctan(y/x)#

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

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

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

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

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

# \ \ \ \ \ \ = -x - 2y arctan(y/x)#

And now if we differentiate again, this time wrt #x# we get:

# (partial ^2u)/(partial x partial y) = -1 - 2y (1/(1+(y/x)^2) (partial)/(partial x)arctan(y/x))#

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

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

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

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