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Пошаговое объяснение:
Djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker djoker
Поскольку (c*f(x))' = c*f(x)', то полученную производную, домножим затем на -1.
Представим выражение как:
y=-1
Решение:
(sin(a-x)*cos(x+a))' = (cos(x+a))'(sin(a-x))+(cos(x+a))*(sin(a-x))' = (-sin(x+a))*(sin(a-x))+(cos(x+a))*(-cos(a-x))
Здесь:
(cos(x+a))' = (cos(x+a))'(x+a)' = -sin(x+a)
(x+a)' = (a)' + (x)' = 0 + 1 = 1
(sin(a-x))' = (sin(a-x))'(a-x)' = -cos(a-x)
(a-x)' = (a)' + (-x)' = 0 + (-1) = -1
ответ:
sin(a-x)*sin(x+a)+cos(a-x)*cos(x+a)
При вычислении были использованы следующие правила дифференцирования:
(xa)' = axa-1
(a)' = 0
(uv)' = u'v + uv'
(f(g(x)))' = f(x)'*g(x)'