derivative f(x)=(2-ln(x))/(2xsqrt(x))

解答

导数

f (x) = \frac{2 - ln ( x )}{2 x x}

\frac{- 8 + 3 ln ( x )}{4 x ^{\frac{5}{2}}}

求解步骤

\frac{d}{d x} (\frac{2 - ln ( x )}{2 x x})

将常数提出:

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

= \frac{1}{2} \frac{d}{d x} (\frac{2 - ln ( x )}{x 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} ( 2 - ln ( x ) ) x x - \frac{d}{d x} ( x x ) ( 2 - ln ( x ) )}{( x x ) ^{2}}

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

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

= \frac{1}{2} \cdot \frac{( - \frac{1}{x} ) x x - \frac{3 x}{2} ( 2 - ln ( x ) )}{( x x ) ^{2}}

化简

\frac{1}{2} \cdot \frac{( - \frac{1}{x} ) x x - \frac{3 x}{2} ( 2 - ln ( x ) )}{( x x ) ^{2}} : \frac{- 8 + 3 ln ( x )}{4 x ^{\frac{5}{2}}}

= \frac{- 8 + 3 ln ( x )}{4 x ^{\frac{5}{2}}}

\frac{\partial}{\partial y} (- 4 x^{3} + 4 x)

\int_{0}^{\frac{π}{28}} tan (14 x) d x

d er i v a t i v e \frac{m x}{b x + c}

\frac{\partial}{\partial x} (2 + 3 x y)

n = 0 \sum \infty (4 x^{2})^{n}