Для функции f(x)=3sin^x вычислите f'(-п/4).
sin в степени?
да
f'(х)=3cosx.
f'(-π/4)=3cosx=3cos (-π/4)=3(-√2/2)=-3√2/2
F'(х) = 3 cosx f''(х) = - 3 sinx f''(-π/4) = - 3 sin(-π/4) = -3 * (-√2)/2 = 1,5√2