(\partial)/(\partial t)((x-ct)^4+(x+ct)^4)

解答

\frac{\partial}{\partial t} ((x - c t)^{4} + (x + c t)^{4})

8 c^{4} t^{3} + 24 c^{2} x^{2} t

求解步骤

\frac{\partial}{\partial t} ((x - c t)^{4} + (x + c t)^{4})

化简

(x - c t)^{4} + (x + c t)^{4} : 2 x^{4} + 12 c^{2} x^{2} t^{2} + 2 c^{4} t^{4}

= \frac{\partial}{\partial t} (2 x^{4} + 12 c^{2} x^{2} t^{2} + 2 c^{4} t^{4})

将

c, x

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial t} (2 x^{4}) + \frac{\partial}{\partial t} (12 c^{2} x^{2} t^{2}) + \frac{\partial}{\partial t} (2 c^{4} t^{4})

\frac{\partial}{\partial t} (2 x^{4}) = 0

\frac{\partial}{\partial t} (12 c^{2} x^{2} t^{2}) = 24 c^{2} x^{2} t

\frac{\partial}{\partial t} (2 c^{4} t^{4}) = 8 c^{4} t^{3}

= 0 + 24 c^{2} x^{2} t + 8 c^{4} t^{3}

化简

= 8 c^{4} t^{3} + 24 c^{2} x^{2} t

\int \frac{( 4 x + 2 )}{x ^{3} - x ^{2} + 2 x} d x

y^{^{'}} = 2 t y^{2}

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

\int - e^{- x} x d x

\int_{0}^{10} f (x) d x