derivative (sqrt(x))/(x^3+1)

Lösung

ableitung von

\frac{x}{x ^{3} + 1}

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

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

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

10 x y y^{^{'}} = 1 - y^{2}

\frac{d}{d x} (6 x - 1)

\int_{- 2}^{2} x^{4} - 8 x^{2} + 16 d x

n = 0 \sum \infty \frac{2 n}{1 + n ^{4}}