derivative (1-4t)/(3+t)

解答

导数

\frac{1 - 4 t}{3 + t}

- \frac{13}{( 3 + t ) ^{2}}

求解步骤

\frac{d}{d t} (\frac{1 - 4 t}{3 + t})

使用微分除法定则:

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

= \frac{\frac{d}{d t} ( 1 - 4 t ) ( 3 + t ) - \frac{d}{d t} ( 3 + t ) ( 1 - 4 t )}{( 3 + t ) ^{2}}

\frac{d}{d t} (1 - 4 t) = - 4

\frac{d}{d t} (3 + t) = 1

= \frac{( - 4 ) ( 3 + t ) - 1 \cdot ( 1 - 4 t )}{( 3 + t ) ^{2}}

化简

\frac{( - 4 ) ( 3 + t ) - 1 \cdot ( 1 - 4 t )}{( 3 + t ) ^{2}} : - \frac{13}{( 3 + t ) ^{2}}

= - \frac{13}{( 3 + t ) ^{2}}

\int \frac{( 2 z - 1 )}{2 z ^{2} - 6 z + 5} d z

(9 x + 1) y^{2} \frac{d y}{d x} + 4 x^{2} + 3 y^{3} = 0

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d er i v a t i v e tan^{2} (6 θ)

in v erse l a pl a ce \frac{1}{s ^{3} ( s - 1 )}