Question

Seja f ( t ) = t cos t. Encontre f```( 0 ).

Answer 1

$\frac{df(t)}{dt}~=~f'(t)~=~(t)'~.~cos(t)~+~t~.~(cos(t))'\\\\\\\frac{df(t)}{dt}~=~f'(t)~=~1~.~cos(t)~+~t~.~-sen(t)\\\\\\\boxed{\frac{df(t)}{dt}~=~f'(t)~=~cos(t)~-~t.sen(t)}\\\\\\\\\\\frac{df(t)}{dt^2}~=~f''(t)~=~(~cos(t)~-~t.sen(t)~)'\\\\\\\frac{df(t)}{dt^2}~=~f''(t)~=~(cos(t))'~-~\left(~(t)'~.~sen(t)~+~t~.~(sen(t))'~\right)\\\\\\\frac{df(t)}{dt^2}~=~f''(t)~=~-sen(t)~-~\left(~1~.~sen(t)~+~t~.~cos(t)~\right)\\\\\\\frac{df(t)}{dt^2}~=~f''(t)~=~-sen(t)~-~\left(~sen(t)~+~t.cos(t)~\right)$

$\frac{df(t)}{dt^2}~=~f''(t)~=~-sen(t)~-~sen(t)~-~t.cos(t)~\right)\\\\\\\boxed{\frac{df(t)}{dt^2}~=~f''(t)~=~-2sen(t)~-~t.cos(t)}\\\\\\\\\\\frac{df(t)}{dt^3}~=~f'''(t)~=~(~-2sen(t)~-~t.cos(t)~)'\\\\\\\frac{df(t)}{dt^3}~=~f'''(t)~=~(~-2sen(t)~)'~-~(~t.cos(t)~)'\\\\\\\frac{df(t)}{dt^3}~=~f'''(t)~=~-2cos(t)~-~(~(t)'~.~cos(t)~+~t~.~(cos(t))'~)\\\\\\\frac{df(t)}{dt^3}~=~f'''(t)~=~-2cos(t)~-~(~1~.~cos(t)~+~t~.~-sen(t)~)\\\\\\\frac{df(t)}{dt^3}~=~f'''(t)~=~-2cos(t)~-~(~cos(t)~-~t.sen(t)~)$

$\frac{df(t)}{dt^3}~=~f'''(t)~=~-2cos(t)~-~cos(t)~+~t.sen(t)\\\\\\\boxed{\frac{df(t)}{dt^3}~=~f'''(t)~=~-3cos(t)~+~t.sen(t)}$

Calculando f'''(0), temos:

$f'''(0)~=~-3cos(0)~+~0.sen(0)\\\\\\f'''(0)~=~-3~.~1~+~0\\\\\\\boxed{f'''(0)~=~-3}$