(\partial)/(\partial y)(-1/(x^2y))

解答

\frac{\partial}{\partial y} (- \frac{1}{x ^{2} y})

\frac{1}{x ^{2} y ^{2}}

求解步骤

\frac{\partial}{\partial y} (- \frac{1}{x ^{2} y})

将

x

视为常数

将常数提出:

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

= - \frac{1}{x ^{2}} \frac{\partial}{\partial y} (\frac{1}{y})

使用指数法则:

\frac{1}{a} = a^{- 1}

= - \frac{1}{x ^{2}} \frac{\partial}{\partial y} (y^{- 1})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= - \frac{1}{x ^{2}} (- 1 \cdot y^{- 1 - 1})

化简

- \frac{1}{x ^{2}} (- 1 \cdot y^{- 1 - 1}) : \frac{1}{x ^{2} y ^{2}}

= \frac{1}{x ^{2} y ^{2}}

\int cos (5 x + 7) d x

y^{^{'}} = 3 - 6 x + y - 2 x y

\int sin (3 x) + cos (5 x) d x

in v erse l a pl a ce \frac{3 s + 3}{s ^{2} - 9}

\frac{\partial}{\partial x} (2 (x + c))