What is the derivative of the inverse of dot (inner) product? $\frac{\partial}{\partial t}\left(\langle A,\;A\rangle\right)^{-1}=?$ What is the derivative of the inverse of dot (inner) product?

$$\frac{\partial}{\partial t}\left(\langle A,\;A\rangle\right)^{-1}=?$$

where $A$ is a vector.
 A: If $A(t)=(a_1(t),...,a_n(t)$, then $\left(<A(t),\;A(t)>\right)^{-1}=\frac{1}{\sum_{j=1}^na_j(t)^2}$.
Can you proceed (quotient rule) ?
A: Let $(\langle A, A\rangle)^{-1}:=b\in\mathbb{K}$ where $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ (a scalar field). Now we need to find $\,db/\,dt$. Notice that $b\langle A, A\rangle=1$ therefore  by product rule of differentiation we get
$$\frac{db}{dt}\langle A, A\rangle+b\frac{d}{dt}\langle A, A\rangle=0\Rightarrow\frac{db}{dt}=-b\frac{\frac{d}{dt}\langle A, A\rangle}{\langle A, A\rangle}=-b\frac{\langle dA/dt, A\rangle+\langle A, dA/dt\rangle}{\langle A, A\rangle}\\\Rightarrow \frac{db}{dt}=-2b\frac{\Re{\langle dA/dt, A\rangle}}{\langle A, A\rangle}$$
where we have used $\langle dA/dt, A\rangle+\langle A, dA/dt\rangle=\langle dA/dt, A\rangle+\overline{\langle dA/dt, A\rangle}=2\Re{\langle dA/dt, A\rangle}$. Throughout we are assuming that $A\neq0$ for the inverse to exist in the first place.
A: Since $\langle A,A\rangle’=2\langle A’,A\rangle$ the   derivative in question is 
$$\frac{-1}{\langle A,A\rangle^2}2\langle A',A\rangle.$$
