derivative of (2(Ae^x+Bx+C))

解答

\frac{d}{d x} ((2) (A e^{x} + B x + C))

2 (e^{x} A + B)

求解步骤

\frac{d}{d x} ((2) (A e^{x} + B x + C))

将

A, B, C

视为常数

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (A e^{x} + B x + C)

使用微分加减法定则:

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

= 2 (\frac{d}{d x} (A e^{x}) + \frac{d}{d x} (B x) + \frac{d C}{d x})

\frac{d}{d x} (A e^{x}) = A e^{x}

\frac{d}{d x} (B x) = B

\frac{d C}{d x} = 0

= 2 (A e^{x} + B + 0)

化简

= 2 (e^{x} A + B)

e x p an d lo g_{10} (3) (x - 2)

s im pl i f y (8 x + 9) (4 x + 5)

e x p an d f (6 x + 20)^{22} d x

e x p an d 2 (e^{x} cos (x)) (e^{x} cos (x) - e^{x} sin (x))

e x p an d \frac{1}{( x - 4 ) ( x ^{2} + 1 )}