derivative of (2x^3+4/(x^2+1))

Lösung

\frac{d}{d x} (\frac{2 x ^{3} + 4}{x ^{2} + 1})

\frac{2 x ^{4} + 6 x ^{2} - 8 x}{( x ^{2} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{3} + 4}{x ^{2} + 1})

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

6 x^{2} (x^{2} + 1) - 2 x (2 x^{3} + 4) : 2 x^{4} + 6 x^{2} - 8 x

= \frac{2 x ^{4} + 6 x ^{2} - 8 x}{( x ^{2} + 1 ) ^{2}}

\int \frac{cos ( x )}{2 + sin ( x )} d x

\int r (cos (θ)) z^{3}

im pl i c i t \frac{d y}{d x}, sin (x y) = 5 x^{4} y^{2} + 6

x \to 1 lim (x e^{- x})

\frac{d}{d x} (\frac{16}{x})