@Xingxing:
What is the reason of your answer? How did you get it?
Dear Xingxing,
Unfortunately, I do not have the direct relationship of the variable 'y' with respect to 't'. As a matter of fact, the solution of the above-mentioned derivative will be used as a constraint in an optimization problem and hence, I only have to use y(t) in the problem and not y'(t).
Are you talking here about functional derivatives?
Then you might use $\frac{\delta y(t_1)}{\delta y(t_2} = \delta(t_1 - t_2)$ (beware that the $\delta$ on the left hand side denotes the functional derivative, but on the right hand side it denotes the Dirac Delta "function"). Then the answer to your question would be $a_t \delta(t-1 -t) = 0$. For functional derivatives, compare for instance Wikipedia (see the link below) and textbooks on calculus of variations or theoretical mechanics.
Ответ. Это оператор, который ставит в соответствие функции y функцию a(t)y(t-1).
What about changing the role of a(t) from a parameter to a variable? What will be happened on the result?
Роль a(t) незначительна.Нужно чтобы а(t)y(t) принадлежало соответствующему пространству функций.
In the case that you treat a(t) as functional of y(t), then you would obtain y(t-1) \delta a(t) / \delta y(t). since the functional derivative of y(t-1) with respect to y(t) vanishes, see above. (There is a product rule also for functional derivatives.).
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