(\partial)/(\partial x)((8y)/(9x^2+4y^2+5))

解答

\frac{\partial}{\partial x} (\frac{8 y}{9 x ^{2} + 4 y ^{2} + 5})

- \frac{144 y x}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}}

求解步骤

\frac{\partial}{\partial x} (\frac{8 y}{9 x ^{2} + 4 y ^{2} + 5})

将

y

视为常数

将常数提出:

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

= 8 y \frac{\partial}{\partial x} (\frac{1}{9 x ^{2} + 4 y ^{2} + 5})

使用指数法则:

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

= 8 y \frac{\partial}{\partial x} ((9 x^{2} + 4 y^{2} + 5)^{- 1})

使用连锁律:

- \frac{1}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}} \frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5)

= - \frac{1}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}} \frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5)

\frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5) = 18 x

= 8 y (- \frac{1}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}} \cdot 18 x)

化简

8 y (- \frac{1}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}} \cdot 18 x) : - \frac{144 y x}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}}

= - \frac{144 y x}{( 9 x ^{2} + 4 y ^{2} + 5 ) ^{2}}

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

n = 1 \sum \infty (- 1)^{n} (- 1)^{n}

\frac{d}{d x} (5 x^{l n (x)})

\frac{d y}{d x} y = x^{x}

t an g e n t 8 x^{3} - 18 x^{2} - 6