Bonjour,
(cos(x))' = -sin(x)
(sin(x))' = cos(x)
(u^n)' = nu'u^(n-1) (^ veut dire puissance)
==> f'(x) = -2sin(x)cos(x) - 2cos(x)sin(x) = -4sin(x)cos(x)
Autre méthode :
cos^2(x) = 1 - sin^2(x)
==> f(x) = 1 - 2sin^2(x) = 2cos^2(x) - 1 = cos(2x)
==> f'(x) = -2sin(2x) en utilisant (cos(u))' = -u'sin(u)
