d/(dp)((p-1)/(p^2+4))

解答

\frac{d}{d p} (\frac{p - 1}{p ^{2} + 4})

\frac{- p ^{2} + 2 p + 4}{( p ^{2} + 4 ) ^{2}}

求解步骤

\frac{d}{d p} (\frac{p - 1}{p ^{2} + 4})

使用微分除法定则:

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

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

\frac{d}{d p} (p - 1) = 1

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

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

化简

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

= \frac{- p ^{2} + 2 p + 4}{( p ^{2} + 4 ) ^{2}}

\int \frac{e ^{x}}{1 + e ^{x} + e ^{2 x}} d x

\int 6 x - 7 d x

\frac{\partial}{\partial x} (- e^{- o} \cdot cos (o))

\frac{\partial}{\partial y} (ln (x - 3 y))

d er i v a t i v e - \frac{1}{9 x ^{\frac{10}{9}}}