derivative of (8x/(x^2+3))

解答

\frac{d}{d x} (\frac{8 x}{x ^{2} + 3})

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

求解步骤

\frac{d}{d x} (\frac{8 x}{x ^{2} + 3})

将常数提出:

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

= 8 \frac{d}{d x} (\frac{x}{x ^{2} + 3})

使用微分除法定则:

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

= 8 \cdot \frac{\frac{d x}{d x} ( x ^{2} + 3 ) - \frac{d}{d x} ( x ^{2} + 3 ) x}{( x ^{2} + 3 ) ^{2}}

\frac{d x}{d x} = 1

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

= 8 \cdot \frac{1 \cdot ( x ^{2} + 3 ) - 2 xx}{( x ^{2} + 3 ) ^{2}}

化简

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

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

\frac{d}{d x} [f (x^{2})] = x^{5}

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

\int \frac{1}{x x + 1} d x

d er i v a t i v e y = ln (\frac{4}{x ^{3}})

\frac{\partial}{\partial x} (x cos (\frac{1}{x ^{2} + y ^{2}}))