derivative of (ax^2+bx+c*e^{-x/4})

解答

\frac{d}{d x} ((a x^{2} + b x + c) \cdot e^{- \frac{x}{4}})

2 a e^{- \frac{x}{4}} x + b e^{- \frac{x}{4}} + \frac{- a e ^{- \frac{x}{4}} x ^{2} - b e ^{- \frac{x}{4}} x - c e ^{- \frac{x}{4}}}{4}

求解步骤

\frac{d}{d x} ((a x^{2} + b x + c) e^{- \frac{x}{4}})

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = a x^{2} + b x + c, g = e^{- \frac{x}{4}}

= \frac{d}{d x} (a x^{2} + b x + c) e^{- \frac{x}{4}} + \frac{d}{d x} (e^{- \frac{x}{4}}) (a x^{2} + b x + c)

\frac{d}{d x} (a x^{2} + b x + c) = 2 a x + b

\frac{d}{d x} (e^{- \frac{x}{4}}) = - \frac{1}{4} e^{- \frac{x}{4}}

= (2 a x + b) e^{- \frac{x}{4}} + (- \frac{1}{4} e^{- \frac{x}{4}}) (a x^{2} + b x + c)

化简

(2 a x + b) e^{- \frac{x}{4}} + (- \frac{1}{4} e^{- \frac{x}{4}}) (a x^{2} + b x + c) : 2 a e^{- \frac{x}{4}} x + b e^{- \frac{x}{4}} + \frac{- a e ^{- \frac{x}{4}} x ^{2} - b e ^{- \frac{x}{4}} x - c e ^{- \frac{x}{4}}}{4}

= 2 a e^{- \frac{x}{4}} x + b e^{- \frac{x}{4}} + \frac{- a e ^{- \frac{x}{4}} x ^{2} - b e ^{- \frac{x}{4}} x - c e ^{- \frac{x}{4}}}{4}

x \to - \infty lim (1 + \frac{1}{x})

t an g e n t 3 x, at x = 16

\int x \cdot sin (2 \cdot x) d x

d er i v a t i v e 450 (1 + \frac{3 t}{29 + t ^{2}}), 4

x \to 0 lim (\frac{( x + 4 )}{2})