derivative e^tcos(t)-e^tsin(t)

解答

导数

e^{t} cos (t) - e^{t} sin (t)

- 2 e^{t} sin (t)

求解步骤

\frac{d}{d t} (e^{t} cos (t) - e^{t} sin (t))

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (e^{t} cos (t)) - \frac{d}{d t} (e^{t} sin (t))

\frac{d}{d t} (e^{t} cos (t)) = e^{t} cos (t) - e^{t} sin (t)

\frac{d}{d t} (e^{t} sin (t)) = e^{t} sin (t) + cos (t) e^{t}

= e^{t} cos (t) - e^{t} sin (t) - (e^{t} sin (t) + cos (t) e^{t})

化简

e^{t} cos (t) - e^{t} sin (t) - (e^{t} sin (t) + cos (t) e^{t}) : - 2 e^{t} sin (t)

= - 2 e^{t} sin (t)

\int \frac{1 - x}{x ^{2} + 1} d x

\frac{\partial}{\partial x} (ln (x + y))

y^{^{'}} = e^{(y - x)}

(\frac{q ^{4}}{4} + \frac{2 q ^{3}}{3} - \frac{11 q ^{2}}{2} + 12 q)^{^{'}}

4 (1 + x^{2}) \frac{d y}{d x} = 2 x y (y^{4} - 1)