I have a question on taking the time derivative (i.e. w.r.t. t variable in the following function) of a quadratic function $V=x(t)^TPx(t)$. $t\in R^+$ has the physical meaning of time. $x(t)\in R^n$, $P\in R^{n\times n}$ is a positive definite matrix. x(t) has its own linear dynamics:$\dot{x}(t)=Ax(t),A\in R^{n\times n}$. I ran in to this problem when I was reading materials of Lyapunov function of linear systems, and from the materials I knew the answer should be $\dot{V}=x^T(A^TP+PA)x$.
The above correct answer can be obtained by applying the product rule:
$\dot{V}=\dot{x}^TPx+x^TP\dot{x}=(Ax)^TPx+x^TP(Ax)=x^T(A^TP+PA)x$.
But when I was deriving it, I use the chain rule, and got different answer:
$\dot{V}=\frac{\partial V}{\partial x}\dot{x}=\frac{\partial x^TPx}{\partial x}\dot{x}=x^T(P+P^T)(Ax)=x^T(PA+P^TA)x=2x^TPAx$, the last step was becasue of the symmetric property of $P$ being positive definite.
Was there anything wrong I did with applying the chain rule? or should product rule being applied before chain rule? Thank you for your answer in advance!
Best Answer
The two derivations agree.
Note first that $\mathbf x^TM\mathbf x=\mathbf x^TM^T\mathbf x$ for any $M$ regardless of symmetry or definiteness. Applying the chain rule, $${dV\over dt}={dV\over d\mathbf x}{d\mathbf x\over dt}=\mathbf x^T(P+P^T)A\mathbf x=x^T(PA+P^TA)\mathbf x=\mathbf x^T(PA+A^TP)\mathbf x.$$ In the last step of your derivation, you replaced $P^T$ with $P$ instead of replacing $P^TA$ with $(P^TA)^T=A^TP$.
Applying the product rule instead (which is a special case of the chain rule), $$\begin{align}{dV\over dt}&=\mathbf x^T{d(P\mathbf x)\over dt}+{d(\mathbf x^T)\over dt}P\mathbf x\\&=\mathbf x^TP{d\mathbf x\over dt}+\left({d\mathbf x\over dt}\right)^TP\mathbf x\\&=\mathbf x^TPA\mathbf x+(A\mathbf x)^TP\mathbf x\\&=\mathbf x^T(PA+A^TP)\mathbf x.\end{align}$$ Note that this derivation itself required two uses of the chain rule.
If in addition $P$ is symmetric, then this expression can be further simplified to $2\mathbf x^TPA\mathbf x$, as you found.