 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017/5/solution

Let

G (s) = \int_{0}^{s} e^{t^{2}} d t

G^{'} (s) = e^{s^{2}}

. Applying the chain rule to

F (x, y) = G (x y^{2})

gives

F_{x} = y^{2} e^{x^{2} y^{4}}, F_{y} = 2 x y e^{x^{2} y^{4}} .

No elementary antiderivative of

e^{t^{2}}

is needed: differentiation of the variable upper limit supplies the integrand directly.

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