derivative of 1/2 x^2-x+1/6 x^3

解答

\frac{d}{d x} (\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3})

x - 1 + \frac{x ^{2}}{2}

求解步骤

\frac{d}{d x} (\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3})

化简

\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3} : \frac{x ^{2}}{2} - x + \frac{x ^{3}}{6}

= \frac{d}{d x} (\frac{x ^{2}}{2} - x + \frac{x ^{3}}{6})

使用微分加减法定则:

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

= \frac{d}{d x} (\frac{x ^{2}}{2}) - \frac{d x}{d x} + \frac{d}{d x} (\frac{x ^{3}}{6})

\frac{d}{d x} (\frac{x ^{2}}{2}) = x

\frac{d x}{d x} = 1

\frac{d}{d x} (\frac{x ^{3}}{6}) = \frac{x ^{2}}{2}

= x - 1 + \frac{x ^{2}}{2}

\int x^{2} \cdot x d x

\int \frac{x}{( x + 1 ) ( x + 2 ) ( x + 3 ) ^{2}} d x

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

\int (3 - x) 7^{(3 - x)^{2}} d x

y^{^{'}} = \frac{7 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}