derivative of (kx+a/(bx+c))

解答

\frac{d}{d x} (\frac{k x + a}{b x + c})

\frac{- ab + c k}{( b x + c ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{k x + a}{b x + c})

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} ( k x + a ) ( b x + c ) - \frac{d}{d x} ( b x + c ) ( k x + a )}{( b x + c ) ^{2}}

\frac{d}{d x} (k x + a) = k

\frac{d}{d x} (b x + c) = b

= \frac{k ( b x + c ) - b ( k x + a )}{( b x + c ) ^{2}}

乘开

k (b x + c) - b (k x + a) : - ab + c k

= \frac{- ab + c k}{( b x + c ) ^{2}}

\int ln (2 x + 1) d x

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

n = 0 \sum \infty 2 + 0.8 6^{n}

x \to 0 lim (sin (5 x) csc (9 x))

in v erse l a pl a ce \frac{( s + 1 ) ^{3}}{s ^{4}}