I noticed this notation while going through a tutorial on matrix calculus: $$ \frac{\partial x^TAx}{\partial xx^T} = \frac{\partial}{\partial x}\left( \frac{\partial x^TAx}{\partial x} \right) = A^T+A $$

I'm not sure how the first equality is established (is the partial derivative of $x^TAx$ with respect to $xx^T$ actually equal to the second partial? if so can somebody plz show me). If the first term is simply a shorthand notation, wouldn't $ \frac{\partial^2 x^TAx}{\partial x\partial x^T} $ or $ \frac{\partial^2 x^TAx}{\partial x^2} $make more sense?

I'm aware of the notation $ \frac{\partial^2 f}{\partial x^2}$ in ordinary calculus, so I don't understand why the numerator and denominator of the shorthand notation above each uses only a single partial symbol...

  • $\begingroup$ After seeing more matrix calculus in action, I came to the conclusion that this shorthand is unfortunately just another abuse of notation. But I'd like to find out what $\frac{\partial x^TAx}{\partial xx^T}$ literally evaluates to $\endgroup$ – Yibo Yang Jul 4 '16 at 13:10

To answer the question in your comment, define a new variable $$M=xx^T$$Then use this new variable and the Frobenius Inner Product to write the function and its differential as $$\eqalign{ f &= A:M \cr df &= A:dM \cr }$$ Since $df=\big(\frac{\partial f}{\partial M}:dM\big),\,$ the gradient must be $$\eqalign{ \frac{\partial f}{\partial M} &= \frac{\partial f}{\partial (xx^T)} &= A \cr }$$ So that's what the expression evaluates to, if taken literally.

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