(\partial)/(\partial x)(1/(2sqrt(1+x+y)))

解答

\frac{\partial}{\partial x} (\frac{1}{2 1 + x + y})

- \frac{1}{4 ( 1 + x + y ) ^{\frac{3}{2}}}

求解步骤

\frac{\partial}{\partial x} (\frac{1}{2 1 + x + y})

将

y

视为常数

将常数提出:

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

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

使用指数法则:

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

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

使用连锁律:

- \frac{1}{2 ( 1 + x + y ) ^{\frac{3}{2}}} \frac{\partial}{\partial x} (1 + x + y)

= - \frac{1}{2 ( 1 + x + y ) ^{\frac{3}{2}}} \frac{\partial}{\partial x} (1 + x + y)

\frac{\partial}{\partial x} (1 + x + y) = 1

= \frac{1}{2} (- \frac{1}{2 ( 1 + x + y ) ^{\frac{3}{2}}} \cdot 1)

化简

\frac{1}{2} (- \frac{1}{2 ( 1 + x + y ) ^{\frac{3}{2}}} \cdot 1) : - \frac{1}{4 ( 1 + x + y ) ^{\frac{3}{2}}}

= - \frac{1}{4 ( 1 + x + y ) ^{\frac{3}{2}}}

\frac{d y}{d x} = \frac{3 x + x y ^{2}}{y + x ^{2} y}

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

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

d er i v a t i v e \frac{2 x ^{2} + 4 x + 6}{x}

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