d/(dy)(x/(y^2+x^2))

解答

\frac{d}{d y} (\frac{x}{y ^{2} + x ^{2}})

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

求解步骤

\frac{d}{d y} (\frac{x}{y ^{2} + x ^{2}})

将

x

视为常数

将常数提出:

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

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

使用指数法则:

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

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

使用连锁律:

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

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

\frac{d}{d y} (y^{2} + x^{2}) = 2 y

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

化简

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

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 4

\int \frac{2 + x}{4 x ^{2} - 4 x - 3} d x

d er i v a t i v e f (x) = ln (x - 1)

\int (12 x^{2} - 8 x + 2) d x

\frac{d y}{d x} = \frac{( x - 1 )}{y}