# Vector derivation of $x^Tx$

Let $x \in \mathbb{R}^n$

What is

$$\frac{\partial}{\partial x} [ x^Tx ]$$

My guess is: $\frac{\partial}{\partial x} [ x^Tx ] = 0$, because $[x^Tx] \in \mathbb{R}^1$, hence a real number as is interpreted as scalar in this derivation.

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i think norm of any vector is always a positive real number so it is constant and derivative of constant function is zero. –  Kns Apr 16 '12 at 9:16
@Kunjanshah: Do you think every function $\Bbb R^n\to \Bbb R$ is constant? –  anon Apr 16 '12 at 9:25

Let $u:\mathbb R^n\to\mathbb R$, $x\mapsto u(x)=x^Tx$. There exists a linear application $\ell_x:\mathbb R^n\to\mathbb R$, called the gradient of $u$ at $x$, such that $u(x+z)=u(x)+\ell_x(z)+o(\|z\|)$ when $z\to0$. To compute $\ell_x$, note that $$u(x+z)=(x+z)^T(x+z)=x^Tx+z^Tx+x^Tz+z^Tz=x^Tx+2x^Tz+o(\|z\|),$$ hence $$\ell_x(z)=2x^Tz.$$ Every linear form $\ell$ on $\mathbb R^n$ has the form $\ell:z\mapsto w^Tz$ for some $w$ in $\mathbb R^n$ hence one often identifies $\ell$ with $w$ (technically, this is identifying the dual of $\mathbb R^n$ with $\mathbb R^n$). In the present case, one identifies the gradient $\ell_x$ of $u$ at $x$ ( a linear application from $\mathbb R^n$ to $\mathbb R$) with the vector $2x$ (an element of $\mathbb R^n$), and one often writes $$(\text{grad}\ u)(x)=2x.$$
Thanks for this specific answer, but I am afraid this does not help me. Is this a counter example, why do you introduce new variables? What is $\frac{\partial}{\partial x} [x^Tx]$ after all? –  Mahoni Apr 16 '12 at 9:40
Nothing specific here, please read again: the object you call $\frac{\partial}{\partial x}(x^Tx)$ (using a notation I cannot recommend) is $(\text{grad}\ u)(x)$, that is, $2x$. –  Did Apr 16 '12 at 9:45
This clarifies things a lot and what about $(grad\ x) (x^T x)$? –  Mahoni Apr 16 '12 at 9:53