derivative f(x)=(x^2)/(x^3+1)

Lösung

ableitung von

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

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

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

\int_{2}^{7} \frac{1}{x - 2} d x

d er i v a t i v e f (x) = e^{4 x s i n (2 x)}

\frac{\partial}{\partial x} (\frac{e ^{x}}{u ( x ) + x ^{2}})

y^{^{''}} - \frac{9}{x} y^{^{'}} + \frac{34}{x ^{2}} y = 0

\frac{d y}{d x} = 5 y + 2