Take the function $f : \mathbb{R}^{+} \times \mathbb{R} \to \mathbb{R}$ defined by $$f(t,x) = \mathbb{1}_{[1-t,2-t]}(x) $$ I am trying to figure out what $\partial_{t}f$ should be in the distributional sense.
My idea is to use the definition of the derivative $$\partial_{t}f = \lim_{h \to 0} \frac{f(t+h,x) - f(t,x)}{h}. $$ using that I got $\partial_{t}f = \delta(x+t-1) - \delta(x+t-2)$. I am not sure if my answer is correct since the function is a multivariable function and I know that the derivative is often defined differently to what I gave above.
If we were to find $\partial_{x}f$ then it would be $\delta(x+t-1) - \delta(x+t-2)$ and I am fine with this. We can just fix time and treat it like a single variable function. But when time is involved I become unsure since we can't just fix $x$ and treat it as a single variable function..
Letting $H(t)=1_{\{t\ge 0\}}$ denote the Heaviside function, the fundamental relation is $$ \frac{d H}{dt}(t)=\delta(t).$$ Now, $$ 1_{[1-t, 2-t]}(x)=H(t+x-1)H(-t-x+2).$$ Differentiating we get $$ \begin{split} \frac{d}{dt}1_{[1-t, 2-t]}(x)&=\delta(t+x-1)H(-t-x+2) - H(t+x-1)\delta(-t-x+2)\\ &=\delta(t+x-1)-\delta(-t-x+2), \end{split}$$ which is the result we wanted.
Remark. In the last identity we have used the property $$ f(t)\delta(t-a)=f(a)\delta(t-a); $$ so, for example, $$\delta(t+x-1)H(-t-x+2)=\delta(t+x-1)H(-1+2)=\delta(t+x-1), $$ since $H(1)=1$.