(\partial)/(\partial y)(zy^2(t+1))

解答

\frac{\partial}{\partial y} (z y^{2} (t + 1))

2 zy (t + 1)

求解步骤

\frac{\partial}{\partial y} (z y^{2} (t + 1))

将

z, t

视为常数

将常数提出:

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

= z (t + 1) \frac{\partial}{\partial y} (y^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= z (t + 1) \cdot 2 y^{2 - 1}

化简

= 2 zy (t + 1)

\int e^{- s i n (x)} cos (x) d x

x \to 0 - lim (1 - \frac{3}{x} + \frac{1}{x ^{2}})

d er i v a t i v e y = (1 + 4 x)^{\frac{3}{2}}

\int x sin (\frac{nπ x}{3}) d x

in v erse l a pl a ce \frac{1}{s ^{2} ( s ^{2} + 1 ^{2} )}