d/(dt)(-4.9t^2+24.5t+6.8)

解答

\frac{d}{d t} (- 4.9 t^{2} + 24.5 t + 6.8)

- 9.8 t + 24.5

求解步骤

\frac{d}{d t} (- 4.9 t^{2} + 24.5 t + 6.8)

使用微分加减法定则:

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

= - \frac{d}{d t} (4.9 t^{2}) + \frac{d}{d t} (24.5 t) + \frac{d}{d t} (6.8)

\frac{d}{d t} (4.9 t^{2}) = 9.8 t

\frac{d}{d t} (24.5 t) = 24.5

\frac{d}{d t} (6.8) = 0

= - 9.8 t + 24.5 + 0

化简

= - 9.8 t + 24.5

\int - \frac{1}{64} cos (8 x) d x

n = 0 \sum \infty - 6 \cdot (\frac{6}{7})^{n}

\frac{d}{d x} (x^{4} + 8 x^{3} + 6 x^{2} - 40 x + 25)

x y^{^{'}} + 2 y = - 3 sin (x)

\int \frac{x ^{3}}{64 - x ^{2}} d x