 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/14/b/ii/solution

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule gives

f_{xx} = 2 f_{u} + 2 y^{2} f_{v} + 4 x^{2} f_{uu} + 8 x^{2} y^{2} f_{uv} + 4 x^{2} y^{4} f_{vv},

f_{yy} = 2 f_{u} + 2 x^{2} f_{v} + 4 y^{2} f_{uu} + 8 x^{2} y^{2} f_{uv} + 4 x^{4} y^{2} f_{vv},

and

f_{x y} = 4 x y f_{uu} + 4 x y (x^{2} + y^{2}) f_{uv} + 4 x^{3} y^{3} f_{vv} + 4 x y f_{v} .

For example, the coefficient of

f_{uv}

u_{x} v_{y} + v_{x} u_{y} = 4 x^{3} y + 4 x y^{3}

, while the final term is

v_{x y} f_{v} = 4 x y f_{v}

f_{uv} = f_{vu}

is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.

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