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Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 2 / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Oxford Undated Prelims Mathematics III resit 2 a
2026-10-25  0 By others on same topic  0 Discussions Create my own version
Put h(s)=sinx(s)−cosy(s). The chain rule gives
h′=x′cosx+y′siny,f′=−h′sinh=−(x′cosx+y′siny)g.
(1)
Apply the product rule and the chain rule once more: g′=h′cosh=h′f, and
h′′=x′′cosx−(x′)2sinx+y′′siny+(y′)2cosy.
(2)
Consequently f′′=−(h′)2f−h′′g, so the requested coefficients are
ab​=−(x′cosx+y′siny)2,=−x′′cosx+(x′)2sinx−y′′siny−(y′)2cosy.​​
(3)
This is the second derivative chain rule along the parametric curve (x(s),y(s)).

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