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Mathematical methods II course of the University of Oxford

Past exam of the mathematical methods II course of the University of Oxford

2016

Use <implicit differentiation> while holding $x$ fixed. Writing $y_z=(\partial y/\partial z)_x$, the constraint gives
$$
2xyy_z+z^2y_z+2yz=0,\qquad y_z=-\frac{2yz}{2xy+z^2}.
$$
The <chain rule> applied to $v=xy^3$ now yields
$$
\boxed{\left(\frac{\partial v}{\partial z}\right)_x=3xy^2y_z=-\frac{6xy^3z}{2xy+z^2},\qquad \left(\frac{\partial v}{\partial z}\right)_{x,(1,1,1)}=-2.}
$$
This local formula requires $2xy+z^2\ne0$, as in the <implicit function theorem>; that denominator equals $3$ at the specified point.

Solution

Past exam of the mathematical methods II course of the University of Oxford / 2016 / 3 / Solution

Past exam of the mathematical methods II course of the University of Oxford / 2016 / 3 / Solution

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Past exam of the mathematical methods II course of the University of Oxford / 2016 / 3 / Solution

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