derivative of (4-3x/(3x^2+x))

Lösung

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

\frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

\frac{d}{d x} (3 x^{2} + x) = 6 x + 1

= \frac{( - 3 ) ( 3 x ^{2} + x ) - ( 6 x + 1 ) ( 4 - 3 x )}{( 3 x ^{2} + x ) ^{2}}

Vereinfache

\frac{( - 3 ) ( 3 x ^{2} + x ) - ( 6 x + 1 ) ( 4 - 3 x )}{( 3 x ^{2} + x ) ^{2}} : \frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

= \frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

\frac{\partial}{\partial x} (\frac{7 x}{y})

\frac{d x}{d y} - 0.06 x = 200

\int \frac{6 x}{1 - x ^{4}} d x

\int_{- 3}^{x} t^{2} e^{t^{2}} d t

\int_{1}^{2} 5^{- θ} d θ