If #f= x^2 y^2+2xy^2#, then show that #(partial ^2f)/(partial x partial y) = (partial ^2f)/(partial y partial x)#. obtain the value of #f_(xy)(1,1)# ?

1 Answer
Steve M
Apr 5, 2018

# f_(xy)(1,1) = 8 #

Explanation:

We have:

# f = x^2y^2+2xy^2 #

We compute the first partial derivatives (by differentiating wrt to specified variable and treating all other variables as constants):

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

# f_y = (partial f)/(partial y) = 2x^2y+4xy #

Next we compute the second partial derivatives:

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

# \ \ \ \ = (partial)/(partial y)( (partial)/(partial x) ) #

# \ \ \ \ = (partial)/(partial y)( 2xy^2+2y^2 ) #

# \ \ \ \ = 4xy + 4y #

And:

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

# \ \ \ \ = (partial)/(partial x)( (partial)/(partial y) ) #

# \ \ \ \ = (partial)/(partial x)( 2x^2y+4xy ) #

# \ \ \ \ = 4xy+4y #

And indeed we can verify that:

# (partial ^2f)/(partial x partial y) = (partial ^2f)/(partial y partial x) \ \ \ # QED

This should come as no surprise due to the continuity of #f# over #x,y in RR#

Using this result we then calculate:

# f_(xy)(1,1) = [4xy + 4y]_(x=1,y=1) #

# \ \ \ \ \ \ \ \ \ \ \ \ = 4(1)(1) + 4(1) #

# \ \ \ \ \ \ \ \ \ \ \ \ = 8 #

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