Matrix derivatives

Question

Can someone please help me with this problem? ?I've already searched for similar examples in some linear algebra textbooks, but I couldn't find any... Thanks a lot!

where $x\in \Bbb R^{n\times 1}$

Answer 1

No need for small $\|h\|$ approximations, just apply the chain rule to the function.

$$f(x) = x(x^Tx)^{-1/2}$$

Differential $$df = dx(x^Tx)^{-1/2} - (1/2)x(2x^Tdx)(x^Tx)^{-3/2}$$

Derivative $${\frac{df}{dx}} = I(x^Tx)(x^Tx)^{-3/2} - xx^T(x^Tx)^{-3/2}$$

Or in terms of nabla $$\nabla_x f(x) =\frac{||x||^2I-xx^T}{||x||^3}= \frac{(x^Tx)I-xx^T}{(x^Tx)^{3/2}}$$

Answer 2

Let $h\in\mathbb R^n$ be a vector, then $$f(x+h)-f(x)=\frac{x+h}{||x+h||}-\frac{x}{||x||}=\frac{||x||h-(||x+h||-||x||)x}{||x||\cdot||x+h||}$$ If $||h||$ is small, we have $$||x+h||=\sqrt{x^Tx+2h^Tx+h^Th}=||x||\sqrt{1+2\frac{h^Tx}{x^Tx}+\frac{h^Th}{x^Tx}}=||x||+\frac{x^Th}{||x||}$$ Thus we have $$\nabla_xf(x)=\frac{||x||I-xx^T/||x||}{||x||^2}=\frac{(x^Tx)I-xx^T}{||x||^3}$$