(\partial)/(\partial t)((x)(x^2+xt+t^2-1))

解答

\frac{\partial}{\partial t} ((x) (x^{2} + x t + t^{2} - 1))

x (x + 2 t)

求解步骤

\frac{\partial}{\partial t} ((x) (x^{2} + x t + t^{2} - 1))

将

x

视为常数

将常数提出:

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

= x \frac{\partial}{\partial t} (x^{2} + x t + t^{2} - 1)

使用微分加减法定则:

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

= x (\frac{\partial}{\partial t} (x^{2}) + \frac{\partial}{\partial t} (x t) + \frac{\partial}{\partial t} (t^{2}) - \frac{\partial}{\partial t} (1))

\frac{\partial}{\partial t} (x^{2}) = 0

\frac{\partial}{\partial t} (x t) = x

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

\frac{\partial}{\partial t} (1) = 0

= x (0 + x + 2 t - 0)

化简

= x (x + 2 t)

e x p an d y^{2} d x + (x y - x^{3}) d y

e x p an d (10 x + 3) 5 - 2 x^{2} - 9 x

e x p an d (1 + cot (θ))^{2}

s im pl i f y (1 - 3 sin (x))^{2}

e x p an d (a + b t) e^{3 t} - \frac{2}{5} e^{- 2 t}