d/(du)(sqrt(u^9+v^5))

解答

\frac{d}{d u} (u^{9} + v^{5})

\frac{9 u ^{8}}{2 u ^{9} + v ^{5}}

求解步骤

\frac{d}{d u} (u^{9} + v^{5})

将

v

视为常数

使用连锁律:

\frac{1}{2 u ^{9} + v ^{5}} \frac{d}{d u} (u^{9} + v^{5})

= \frac{1}{2 u ^{9} + v ^{5}} \frac{d}{d u} (u^{9} + v^{5})

\frac{d}{d u} (u^{9} + v^{5}) = 9 u^{8}

= \frac{1}{2 u ^{9} + v ^{5}} \cdot 9 u^{8}

化简

\frac{1}{2 u ^{9} + v ^{5}} \cdot 9 u^{8} : \frac{9 u ^{8}}{2 u ^{9} + v ^{5}}

= \frac{9 u ^{8}}{2 u ^{9} + v ^{5}}

y^{^{'}} + (x + 1) y = e^{x^{2}} y^{3}

\int (4 x^{2} - 8 x + 1) d x

n = 0 \sum \infty (1 - \frac{1}{n})^{n}

\frac{d x}{d t} = \frac{3}{x}

x \to 0 - lim ((\frac{1}{2})^{\frac{1}{x}})