derivative (2x^4)/(b^2-x^2)

Lösung

ableitung von

\frac{2 x ^{4}}{b ^{2} - x ^{2}}

\frac{2 ( 4 b ^{2} x ^{3} - 2 x ^{5} )}{( b ^{2} - x ^{2} ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{4}}{b ^{2} - x ^{2}})

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

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

= 2 \cdot \frac{\frac{d}{d x} ( x ^{4} ) ( b ^{2} - x ^{2} ) - \frac{d}{d x} ( b ^{2} - x ^{2} ) x ^{4}}{( b ^{2} - x ^{2} ) ^{2}}

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

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

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

Vereinfache

2 \cdot \frac{4 x ^{3} ( b ^{2} - x ^{2} ) - ( - 2 x ) x ^{4}}{( b ^{2} - x ^{2} ) ^{2}} : \frac{2 ( 4 b ^{2} x ^{3} - 2 x ^{5} )}{( b ^{2} - x ^{2} ) ^{2}}

= \frac{2 ( 4 b ^{2} x ^{3} - 2 x ^{5} )}{( b ^{2} - x ^{2} ) ^{2}}

\int \frac{( x ^{2} - x + 24 )}{x ^{3} + 6 x} d x

y = (5 t)^{t}

x \to 0 lim (\frac{2 sin ( x ) - sin ( 2 x )}{sin ( x ) - x cos ( x )})

\int_{0}^{4} 3 t^{4} + 3 t + 2 d t

n = 0 \sum \infty (\frac{x}{3})^{n}