(\partial)/(\partial s)((s^2+4)/(t-3))

解答

\frac{\partial}{\partial s} (\frac{s ^{2} + 4}{t - 3})

\frac{2 s}{t - 3}

求解步骤

\frac{\partial}{\partial s} (\frac{s ^{2} + 4}{t - 3})

将

t

视为常数

将常数提出:

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

= \frac{1}{t - 3} \frac{\partial}{\partial s} (s^{2} + 4)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{t - 3} (\frac{\partial}{\partial s} (s^{2}) + \frac{\partial}{\partial s} (4))

\frac{\partial}{\partial s} (s^{2}) = 2 s

\frac{\partial}{\partial s} (4) = 0

= \frac{1}{t - 3} (2 s + 0)

化简

\frac{1}{t - 3} (2 s + 0) : \frac{2 s}{t - 3}

= \frac{2 s}{t - 3}

x \to 0 - lim (\frac{2 x}{∣ x ∣})

t y^{^{''}} - y^{^{'}} = t^{2} + t, y (1) = 1, y^{^{'}} (1) = 5

s l o p e (3, - 7), (10, - 4)

\frac{d}{d x} (3 8 x^{7} + 9 x^{6})

\int \frac{1}{x ^{2} - 4 x + 4} d x