 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/9/a/solution

F_{x} = y^{2} (1 + 2 x^{2} y) e^{x^{2} y}, F_{y} = x (2 y + x^{2} y^{2}) e^{x^{2} y} .

Differentiating the first expression with respect to

y

yields

e^{x^{2} y} (2 y + 7 x^{2} y^{2} + 2 x^{4} y^{3})

; differentiating the second with respect to

x

gives exactly the same polynomial factor. Thus

F_{x y} = F_{y x} = e^{x^{2} y} (2 y + 7 x^{2} y^{2} + 2 x^{4} y^{3}) .

This also follows from symmetry of second derivatives, since all derivatives of this function are continuous everywhere.

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