derivative f(x)=(x^3)/((x^2+4)^2)

解答

导数

f (x) = \frac{x ^{3}}{( x ^{2} + 4 ) ^{2}}

\frac{x ^{2} ( - x ^{2} + 12 )}{( x ^{2} + 4 ) ^{3}}

求解步骤

\frac{d}{d x} (\frac{x ^{3}}{( x ^{2} + 4 ) ^{2}})

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

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

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

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

化简

\frac{3 x ^{2} ( x ^{2} + 4 ) ^{2} - 4 x ( x ^{2} + 4 ) x ^{3}}{( ( x ^{2} + 4 ) ^{2} ) ^{2}} : \frac{x ^{2} ( - x ^{2} + 12 )}{( x ^{2} + 4 ) ^{3}}

= \frac{x ^{2} ( - x ^{2} + 12 )}{( x ^{2} + 4 ) ^{3}}

y^{^{'}} - \frac{3 y}{x} = x

d er i v a t i v e sin (x^{2}) + \frac{( 1 - x )}{1 + x}

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

k = 0 \sum \infty 5 x^{k}

\frac{d}{d x} (2 x - 9)