derivative of (4+ln(2x)/(2020+x))

解答

\frac{d}{d x} (\frac{4 + ln ( 2 x )}{2020 + x})

\frac{2020 - 3 x - x ln ( 2 x )}{x ( 2020 + x ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{4 + ln ( 2 x )}{2020 + x})

使用微分除法定则:

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

= \frac{\frac{d}{d x} ( 4 + ln ( 2 x ) ) ( 2020 + x ) - \frac{d}{d x} ( 2020 + x ) ( 4 + ln ( 2 x ) )}{( 2020 + x ) ^{2}}

\frac{d}{d x} (4 + ln (2 x)) = \frac{1}{x}

\frac{d}{d x} (2020 + x) = 1

= \frac{\frac{1}{x} ( 2020 + x ) - 1 \cdot ( 4 + ln ( 2 x ) )}{( 2020 + x ) ^{2}}

化简

\frac{\frac{1}{x} ( 2020 + x ) - 1 \cdot ( 4 + ln ( 2 x ) )}{( 2020 + x ) ^{2}} : \frac{2020 - 3 x - x ln ( 2 x )}{x ( 2020 + x ) ^{2}}

= \frac{2020 - 3 x - x ln ( 2 x )}{x ( 2020 + x ) ^{2}}

t a y l or e^{4 x}, 3

d er i v a t i v e 7 arctan (x - 1 + x^{2})

l a pl a ce t r an s f or m 8 e^{- 6 t} cos (4 t + 30)

\int_{- \infty}^{0} \frac{1}{4 - 10 x} d x

t an g e n t f (x) = - 2 x^{3}, at x = 1