Past exam of the options I course of the University of Oxford / 2012 / F / F1 / b

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- Solution b

Solution

For differentiable maps with compatible domains, the multivariable chain rule is

D (h_{2} \circ h_{1}) (x) = D h_{2} (h_{1} (x)) \circ D h_{1} (x) .

In Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

\partial_{j} (h_{2} \circ h_{1})_{ℓ} (x) = \sum_{r = 1}^{m} \partial_{r} (h_{2})_{ℓ} (h_{1} (x)) \partial_{j} (h_{1})_{r} (x) .

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