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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

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

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

\frac{d x}{d x} = 1

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

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

Vereinfache

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

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

\frac{d}{d x} (2^{x} (π x^{2} + cos (x)))

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{7})

in v erse l a pl a ce \frac{1}{s ( s ^{2} - 1 )}

t an g e n t f (x) = \frac{1}{x + 2}, at x = 7

d er i v a t i v e f (x) = \frac{x ^{2}}{x + 5}