How to take derivative about $V(x)=x^{T}Px$?

Question

Consider the quadratic function $V(x(t))=x(t)^{T}Px(t)$, where $P$ is a positive symmetric matrix.

How do I calculate its time-derivative, that is, $$\frac{\mathrm{d}}{\mathrm{d} t}V(x(t))?$$

Answer 1

Using the product rule and abbreviating $\frac{d}{dt}$ with a dot, we have $$ \dot V = \dot x^T P x + x^T P \dot x $$ Each of the terms on the right is a scalar, hence symmetric. So $$ \dot x^T Px = (\dot x^T P x)^T = x^T P^T \dot x = x^T P \dot x $$ Therefore $\dot V = 2x^T P \dot x$.

If you prefer working in indices, write $x = (x^1,x^2,\dots,x^n)$ and $V = x^i p_{ij} x^j$ (we use the Einstein summation notation convention, so that the sigma is understood). Then \begin{align*} \dot V = \dot x^i p_{ij} x^j + x^i p_{ij} \dot x^j \end{align*} Swapping the names of the indices on the first term, and using the fact that $P$ is symmetric, we have $$ \dot V = \dot x^j p_{ji} x^i + x^i p_{ij} \dot x^j = 2 x^i p_{ij} \dot x^j $$

To illustrate in the case $n=2$, try $P = \begin{bmatrix} 2 & 1 \\ 1 & 1\end{bmatrix}$. Then \begin{align*} V &= \begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 2x+y \\ x+y \end{bmatrix} \\&= 2x^2 + xy + yx + y^2 = 2x^2 + 2xy + y^2 \end{align*} Therefore \begin{align*} \dot V &= 4x\dot x + 2(\dot x y + x \dot y) + 2 y\dot y \\&= 2(2x\dot x + x\dot y + y \dot x + y\dot y) \\&= 2 \begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} \dot x \\ \dot y \end{bmatrix} \end{align*}

Commenter Aaron is right that the product rule works for just about any bilinear operation.

Answer 2

Noting $f'$ the derivative of the function $f$,

Let: $f = y \mapsto y^T P y$.

$P$ is symmetric, so we have: $f'(y) = 2 y^T P$ (see here).

So: $V = f \circ x$.

Using this formula: $(h \circ g)'(t) = h'(g(t)).g'(t)$,

$$V' = 2 x^T P x'$$