d/(du)(sqrt(u^2+v^2))

解答

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

\frac{u}{u ^{2} + v ^{2}}

求解步骤

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

将

v

视为常数

使用连锁律:

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

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

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

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

化简

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

= \frac{u}{u ^{2} + v ^{2}}

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{7 ^{n + 1}}

\int 8 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

j o in a e^{v} + \frac{b}{v} + \frac{c}{v ^{2}}

t an g e n t \frac{x}{6 x + 4}

\frac{d}{d x} (3 csc (x) + \frac{5}{x})