derivative 12+2.8sin((2pi)/(365)(t-80))

解答

导数

12 + 2.8 sin (\frac{2 π}{365} (t - 80))

0.04819 \dots cos (0.01721 \dots (t - 80))

求解步骤

\frac{d}{d t} (12 + 2.8 sin (\frac{2 π}{365} (t - 80)))

化简

12 + 2.8 sin (\frac{2 π}{365} (t - 80)) : 12 + 2.8 sin (0.01721 \dots (t - 80))

= \frac{d}{d t} (12 + 2.8 sin (0.01721 \dots (t - 80)))

使用微分加减法定则:

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

= \frac{d}{d t} (12) + \frac{d}{d t} (2.8 sin (0.01721 \dots (t - 80)))

\frac{d}{d t} (12) = 0

\frac{d}{d t} (2.8 sin (0.01721 \dots (t - 80))) = 0.04819 \dots cos (0.01721 \dots (t - 80))

= 0 + 0.04819 \dots cos (0.01721 \dots (t - 80))

化简

= 0.04819 \dots cos (0.01721 \dots (t - 80))

n = 0 \sum \infty \frac{4}{2 ^{n}}

f (x) = cos (x)

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