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

Lösung

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

\frac{- 2 x ^{5} + 2 x}{( x ^{4} + 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{2 x ( x ^{4} + 1 ) - 4 x ^{3} x ^{2}}{( x ^{4} + 1 ) ^{2}} : \frac{- 2 x ^{5} + 2 x}{( x ^{4} + 1 ) ^{2}}

= \frac{- 2 x ^{5} + 2 x}{( x ^{4} + 1 ) ^{2}}

a re a y = 4 - x^{2}, 2 x + y = 4

f (x) = x^{2} sin (x)

\int \frac{2 x + 1}{x ^{2} + x - 1} d x

x \to 3 + lim (\frac{2 x}{x - 3})

\frac{\partial}{\partial x} (e^{- 3 x} sin (2 t - 3 x))