derivative of (7x^3+x/(6sqrt(x)))

解答

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

\frac{35 x ^{2} + 1}{12 x}

求解步骤

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

将常数提出:

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

= \frac{1}{6} \frac{d}{d x} (\frac{7 x ^{3} + x}{x})

使用微分除法定则:

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

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( 7 x ^{3} + x ) x - \frac{d}{d x} ( x ) ( 7 x ^{3} + x )}{( x ) ^{2}}

\frac{d}{d x} (7 x^{3} + x) = 21 x^{2} + 1

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{1}{6} \cdot \frac{( 21 x ^{2} + 1 ) x - \frac{1}{2 x} ( 7 x ^{3} + x )}{( x ) ^{2}}

化简

\frac{1}{6} \cdot \frac{( 21 x ^{2} + 1 ) x - \frac{1}{2 x} ( 7 x ^{3} + x )}{( x ) ^{2}} : \frac{35 x ^{2} + 1}{12 x}

= \frac{35 x ^{2} + 1}{12 x}

\frac{d}{d x} (lo g_{10} (\frac{1}{x}))

y^{^{''}} - 6 y^{^{'}} = 4 x

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

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

d er i v a t i v e f (x) = (3 - x)^{4} (x + 5)^{3}