d/(dt)((x)(a(at^2+bt+c)+b)^{r(x,t)})

解答

\frac{d}{d t} ((x) (a (a t^{2} + b t + c) + b)^{r (x, t)})

x e^{r (x, t) l n (a (a t^{2} + b t + c) + b)} (\frac{d}{d t} (r (x, t)) ln (a (a t^{2} + b t + c) + b) + \frac{a ( 2 a t + b )}{a ( a t ^{2} + b t + c ) + b} r (x, t))

求解步骤

\frac{d}{d t} ((x) (a (a t^{2} + b t + c) + b)^{r (x, t)})

将

a, b, c, x

视为常数

将常数提出:

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

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

使用指数法则:

a^{b} = e^{b l n (a)}

(a (a t^{2} + b t + c) + b)^{r (x, t)} = e^{r (x, t) l n (a (a t^{2} + b t + c) + b)}

= x \frac{d}{d t} (e^{r (x, t) l n (a (a t^{2} + b t + c) + b)})

使用连锁律:

e^{r (x, t) l n (a (a t^{2} + b t + c) + b)} \frac{d}{d t} (r (x, t) ln (a (a t^{2} + b t + c) + b))

= e^{r (x, t) l n (a (a t^{2} + b t + c) + b)} \frac{d}{d t} (r (x, t) ln (a (a t^{2} + b t + c) + b))

\frac{d}{d t} (r (x, t) ln (a (a t^{2} + b t + c) + b)) = \frac{d}{d t} (r (x, t)) ln (a (a t^{2} + b t + c) + b) + \frac{a ( 2 a t + b )}{a ( a t ^{2} + b t + c ) + b} r (x, t)

= x e^{r (x, t) l n (a (a t^{2} + b t + c) + b)} (\frac{d}{d t} (r (x, t)) ln (a (a t^{2} + b t + c) + b) + \frac{a ( 2 a t + b )}{a ( a t ^{2} + b t + c ) + b} r (x, t))

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