(\partial)/(\partial y)((x-3)/(y^2+1))

解答

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

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

求解步骤

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

将

x

视为常数

将常数提出:

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

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

使用指数法则:

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

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

使用连锁律:

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

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

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

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

化简

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

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

x \to - \infty lim (x - 5)

\int_{- 2}^{2} 32 - 8 y^{2} d y

\int_{6}^{8} \frac{46}{( x - 6 ) ^{3}} d x

t an g e n t f (x) = \frac{6 x}{x ^{2} - 3}, (3, 3)

\frac{d y}{d x} = - 4 y + x^{3} - x