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

解答

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

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

求解步骤

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

将常数提出:

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

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

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d t}{d t} = 1

\frac{d}{d t} (2 t^{2} + 1) = 4 t

= 2 \cdot \frac{1 \cdot ( 2 t ^{2} + 1 ) - 4 tt}{( 2 t ^{2} + 1 ) ^{2}}

化简

2 \cdot \frac{1 \cdot ( 2 t ^{2} + 1 ) - 4 tt}{( 2 t ^{2} + 1 ) ^{2}} : \frac{2 ( - 2 t ^{2} + 1 )}{( 2 t ^{2} + 1 ) ^{2}}

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

\int - 3 sec (x) tan (x) - 3 csc (x) cot (x) d x

\int \frac{1}{6 x} d x

(t - sin (t))^{^{'}}

n = 1 \sum \infty \frac{4}{n ln ( 8 n )}

d er i v a t i v e f (x) = 17 x