derivative of 1/((x^2+y^2))

解答

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

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

求解步骤

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

使用指数法则:

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

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

使用连锁律:

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

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

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

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

化简

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

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

\frac{\partial}{\partial x} (e^{7 x y})

\int sin^{3} (x) - sin^{5} (x) d x

d er i v a t i v e f (x) = \frac{x}{x ^{2} - 4}

9 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

y^{^{'}} = y (x y^{5} + 1)