derivative of cos(2x+sin(2x)+1/5 e^x)

解答

\frac{d}{d x} (cos (2 x) + sin (2 x) + \frac{1}{5} e^{x})

- sin (2 x) \cdot 2 + cos (2 x) \cdot 2 + \frac{1}{5} e^{x}

求解步骤

\frac{d}{d x} (cos (2 x) + sin (2 x) + \frac{1}{5} e^{x})

使用微分加减法定则:

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

= \frac{d}{d x} (cos (2 x)) + \frac{d}{d x} (sin (2 x)) + \frac{d}{d x} (\frac{1}{5} e^{x})

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

\frac{d}{d x} (sin (2 x)) = cos (2 x) \cdot 2

\frac{d}{d x} (\frac{1}{5} e^{x}) = \frac{1}{5} e^{x}

= - sin (2 x) \cdot 2 + cos (2 x) \cdot 2 + \frac{1}{5} e^{x}

in v erse l a pl a ce \frac{s}{( s - 3 ) ( s - 4 ) ( s - 12 )}

\frac{\partial}{\partial x} (2 x^{\frac{3}{2}})

a re a y = - x^{2} + 6 x, y = x^{2} - 2 x

d er i v a t i v e \frac{3}{5 x ^{4}}

4 y^{^{''}} + 3 y^{^{'}} + 7 y = 0