derivative f(t)=0.1(400-40t+t^2)

解答

导数

f (t) = 0.1 (400 - 40 t + t^{2})

0.1 (2 t - 40)

求解步骤

\frac{d}{d t} (0.1 (400 - 40 t + t^{2}))

将常数提出:

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

= 0.1 \frac{d}{d t} (400 - 40 t + t^{2})

使用微分加减法定则:

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

= 0.1 (\frac{d}{d t} (400) - \frac{d}{d t} (40 t) + \frac{d}{d t} (t^{2}))

\frac{d}{d t} (400) = 0

\frac{d}{d t} (40 t) = 40

\frac{d}{d t} (t^{2}) = 2 t

= 0.1 (0 - 40 + 2 t)

化简

= 0.1 (2 t - 40)

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