Let $A(t)$ be a symmetric positive definite matrix, thus by Cholesky decomposition, we have $A(t)=L(t)L^T(t)$ where $L(t)$ is lower triangular. Suppose $A(t)$ is differentiable.
I want to write $\frac{d}{dt}L(t)$ as a function of $L(t)$, $L^{-1}(t)$ and $\frac{d}{dt}A(t)$.
I tried $$\frac{d}{dt}A(t)=\frac{d}{dt}L(t)L^T(t) + L(t)\frac{d}{dt}L^T(t).$$ I don't know how to proceed. I know that I need to separate $\frac{d}{dt}L(t)$ out. But it seems a troublesome task. Any hint please? Thank you!
For compactness I'll write $C' := \frac{d}{dt} C(t)$ for any differentiable matrix-valued function.
Hint By the product rule you've found that $$A' = L' L^{\top} + L (L')^{\top} .$$ Multiplying both sides on the left by $L^{-1}$ and on the right by $(L^{\top})^{-1}$ and canceling leaves $$L^{-1} A' (L^{-1})^{\top} = L^{-1} L' + (L')^{\top} (L^{-1})^{\top} = L^{-1} L' + (L^{-1} L')^{\top} .$$
The key observations here are that (1) the first term on the r.h.s. is lower-triangular (it's a product of lower-triangular matrices) and (2) the two terms on the r.h.s. are transposes of one another.
Additional hint If $M$ is lower-triangular, then \begin{align}M + M^{\top} &= \pmatrix{ m_{11} & 0 & \cdots & 0 \\ m_{21} & m_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ m_{n1} & m_{n2} & \cdots & m_{nn} } + \pmatrix{ m_{11} & m_{21} & \cdots & m_{n1} \\ 0 & m_{22} & \cdots & m_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & m_{nn} } \\ &= \pmatrix{ 2 m_{11} & m_{21} & \cdots & m_{n1} \\ m_{21} & 2 m_{22} & \cdots & m_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ m_{n1} & m_{n2} & \cdots & 2 m_{nn} } \end{align} In particular, we can read off all of the entries of $M$ from the entries of $M + M^{\top}$.
Applying this procedure to $M = L^{-1} L'$ recovers $L^{-1} L'$ from $L^{-1} A' (L^{-1})^{\top}$.