derivative of ((x^2-1^3)/((2x+1)^5))

Lösung

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

\frac{2 ( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 5 )}{( 2 x + 1 ) ^{6}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} ((x^{2} - 1)^{3}) = 6 x (x^{2} - 1)^{2}

\frac{d}{d x} ((2 x + 1)^{5}) = 10 (2 x + 1)^{4}

= \frac{6 x ( x ^{2} - 1 ) ^{2} ( 2 x + 1 ) ^{5} - 10 ( 2 x + 1 ) ^{4} ( x ^{2} - 1 ) ^{3}}{( ( 2 x + 1 ) ^{5} ) ^{2}}

Vereinfache

\frac{6 x ( x ^{2} - 1 ) ^{2} ( 2 x + 1 ) ^{5} - 10 ( 2 x + 1 ) ^{4} ( x ^{2} - 1 ) ^{3}}{( ( 2 x + 1 ) ^{5} ) ^{2}} : \frac{2 ( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 5 )}{( 2 x + 1 ) ^{6}}

= \frac{2 ( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 5 )}{( 2 x + 1 ) ^{6}}

\int_{- 2}^{4} (x^{2} - 2 x - 8) d x

x \to \infty lim (\frac{sin ( 17 x )}{x})

\frac{\partial}{\partial x} (sin (x^{2} + x y))

\frac{d}{d x} (\frac{x cos ( x )}{1 + e ^{x}})

\frac{\partial}{\partial x} (y^{5} sin (2 x))