derivative f(x)=(4x^3)/(2x^2-5)

解

導関数

f (x) = \frac{4 x ^{3}}{2 x ^{2} - 5}

\frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{4 x ^{3}}{2 x ^{2} - 5})

定数を除去:

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

= 4 \frac{d}{d x} (\frac{x ^{3}}{2 x ^{2} - 5})

商の法則を適用:

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

= 4 \cdot \frac{\frac{d}{d x} ( x ^{3} ) ( 2 x ^{2} - 5 ) - \frac{d}{d x} ( 2 x ^{2} - 5 ) x ^{3}}{( 2 x ^{2} - 5 ) ^{2}}

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

\frac{d}{d x} (2 x^{2} - 5) = 4 x

= 4 \cdot \frac{3 x ^{2} ( 2 x ^{2} - 5 ) - 4 x x ^{3}}{( 2 x ^{2} - 5 ) ^{2}}

簡素化

4 \cdot \frac{3 x ^{2} ( 2 x ^{2} - 5 ) - 4 x x ^{3}}{( 2 x ^{2} - 5 ) ^{2}} : \frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}

= \frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}