 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016/3/solution

x

y_{z} = (\partial y / \partial z)_{x}

, the constraint gives

2 x y y_{z} + z^{2} y_{z} + 2 yz = 0, y_{z} = - \frac{2 yz}{2 x y + z ^{2}} .

The chain rule applied to

v = x y^{3}

now yields

(\frac{\partial v}{\partial z})_{x} = 3 x y^{2} y_{z} = - \frac{6 x y ^{3} z}{2 x y + z ^{2}}, (\frac{\partial v}{\partial z})_{x, (1, 1, 1)} = - 2.

This local formula requires

2 x y + z^{2} \neq = 0

3

at the specified point.

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