I have some doubts regarding the following partial derivative:
\begin{equation} \frac{\partial}{\partial x(t)} \int_{t_0}^{t} h(t-\tau) x(\tau)\, \mathrm{d}\tau. \end{equation}
Assuming that $h(t)$ is the impulse response of a causal and stable linear system, and that $x(t)$ is the system input, I am tempted to do something like: \begin{equation} \frac{\partial}{\partial x(t)} \int_{t_0}^{t} h(t-\tau) x(\tau)\, \mathrm{d}\tau = \int_{t_0}^{t} h(t-\tau)\dot{x}(\tau) \, \mathrm{d}\tau. \end{equation} Because of the linearity of the convolution operator, and because variations in the input ($x(t)$) can be represented as time-varying variations. However, applying Leibnitz I get: \begin{align} \frac{\partial t \partial}{\partial x(t) \partial t} \int_{t_0}^{t} h(t-\tau) x(\tau)\, \mathrm{d}\tau &= \frac{1}{\partial{x}(t)/\partial{t}}\frac{\partial}{\partial t}\int_{t_0}^{t} h(t-\tau)x(\tau) \, \mathrm{d}\tau = \, ... \\ = \frac{1}{\dot{x}(t)} \left[h(0)x(t) + \int_{t_0}^{t} \dot{h}(t-\tau) x(\tau)\, \mathrm{d}\tau \right] &= \frac{1}{\dot{x}(t)} \left[h(0)x(t) + \int_{t_0}^{t} {h}(t-\tau) \dot{x}(\tau)\, \mathrm{d}\tau \right]. \end{align}
I sincerely appreciate any assistance with this.
Cheers!