derivative of ((-x^2-5)/((x^2-5)^2))

Lösung

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

- \frac{2 x ( - x ^{2} - 15 )}{( x ^{2} - 5 ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{( - 2 x ) ( x ^{2} - 5 ) ^{2} - 4 x ( x ^{2} - 5 ) ( - x ^{2} - 5 )}{( ( x ^{2} - 5 ) ^{2} ) ^{2}} : - \frac{2 x ( - x ^{2} - 15 )}{( x ^{2} - 5 ) ^{3}}

= - \frac{2 x ( - x ^{2} - 15 )}{( x ^{2} - 5 ) ^{3}}

l a pl a ce t r an s f or m - 2 e^{3 t}

25 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

\frac{d}{d x} (\frac{e ^{x} ( x ^{2} - 4 x + 5 )}{( 1 - x ) ^{3}})

\int tan (x) d x

\int \frac{1}{x 25 - x ^{2}} d x