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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\int_{0}^{π} sin (m x) e^{n x} d x

\int sin (\frac{x π}{2}) d x

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

l a pl a ce t r an s f or m g (t) = 4 cos (4 t) - 9 sin (4 t) + 2 cos (10 t)

\frac{\partial}{\partial x} (r sin (t) sin (x))