d/(dt)(t^2+1)

解答

\frac{d}{d t} (t^{2} + 1)

2 t

求解步骤

\frac{d}{d t} (t^{2} + 1)

使用微分加减法定则:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (1)

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

\frac{d}{d t} (1) = 0

= 2 t + 0

化简

= 2 t

d er i v a t i v e \frac{x}{x + 2}

5 y^{^{'}} + y = 0

\int_{1}^{7} \frac{x ^{2} - 1}{x} d x

\frac{d}{d t} (t - 1)

d er i v a t i v e f (x) = 645120