derivative of e^{xi}

解答

\frac{d}{d x} (e^{x i})

- sin (x) + i cos (x)

求解步骤

\frac{d}{d x} (e^{x i})

化简

e^{x i} : cos (x) + i sin (x)

= \frac{d}{d x} (cos (x) + i sin (x))

使用微分加减法定则:

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

= \frac{d}{d x} (cos (x)) + \frac{d}{d x} (i sin (x))

\frac{d}{d x} (cos (x)) = - sin (x)

\frac{d}{d x} (i sin (x)) = i cos (x)

= - sin (x) + i cos (x)

\int 3 cos^{2} (x) sin (x) d x

\int (2 t^{2} + 3)^{\frac{4}{3}} t^{3} d t

d er i v a t i v e f (x) = (x^{2} + 3 x - 1) e^{x}

n \to \infty lim (\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}})

x \to 2 lim (3 x - 1)