derivative of (8x^3-2x^2+3/(2x))

解答

\frac{d}{d x} (\frac{8 x ^{3} - 2 x ^{2} + 3}{2 x})

\frac{16 x ^{3} - 2 x ^{2} - 3}{2 x ^{2}}

求解步骤

\frac{d}{d x} (\frac{8 x ^{3} - 2 x ^{2} + 3}{2 x})

将常数提出:

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

= \frac{1}{2} \frac{d}{d x} (\frac{8 x ^{3} - 2 x ^{2} + 3}{x})

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 8 x ^{3} - 2 x ^{2} + 3 ) x - \frac{d x}{d x} ( 8 x ^{3} - 2 x ^{2} + 3 )}{x ^{2}}

\frac{d}{d x} (8 x^{3} - 2 x^{2} + 3) = 24 x^{2} - 4 x

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

= \frac{1}{2} \cdot \frac{( 24 x ^{2} - 4 x ) x - 1 \cdot ( 8 x ^{3} - 2 x ^{2} + 3 )}{x ^{2}}

化简

\frac{1}{2} \cdot \frac{( 24 x ^{2} - 4 x ) x - 1 \cdot ( 8 x ^{3} - 2 x ^{2} + 3 )}{x ^{2}} : \frac{16 x ^{3} - 2 x ^{2} - 3}{2 x ^{2}}

= \frac{16 x ^{3} - 2 x ^{2} - 3}{2 x ^{2}}

\int (y - x^{3}) d x

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

\frac{d y}{d x} + y^{9} x + 4 y = 0

\int_{0}^{2} 4 x + 1 d x

x \to 0 lim (\frac{ln ( 6 x + 1 )}{x})