derivative of (-x^2-9/((x^2-9)^2))

Lösung

\frac{d}{d x} (\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}})

- \frac{2 x ( - x ^{2} - 27 )}{( x ^{2} - 9 ) ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( - x ^{2} - 9 ) ( x ^{2} - 9 ) ^{2} - \frac{d}{d x} ( ( x ^{2} - 9 ) ^{2} ) ( - x ^{2} - 9 )}{( ( x ^{2} - 9 ) ^{2} ) ^{2}}

\frac{d}{d x} (- x^{2} - 9) = - 2 x

\frac{d}{d x} ((x^{2} - 9)^{2}) = 4 x (x^{2} - 9)

= \frac{( - 2 x ) ( x ^{2} - 9 ) ^{2} - 4 x ( x ^{2} - 9 ) ( - x ^{2} - 9 )}{( ( x ^{2} - 9 ) ^{2} ) ^{2}}

Vereinfache

\frac{( - 2 x ) ( x ^{2} - 9 ) ^{2} - 4 x ( x ^{2} - 9 ) ( - x ^{2} - 9 )}{( ( x ^{2} - 9 ) ^{2} ) ^{2}} : - \frac{2 x ( - x ^{2} - 27 )}{( x ^{2} - 9 ) ^{3}}

= - \frac{2 x ( - x ^{2} - 27 )}{( x ^{2} - 9 ) ^{3}}

(x^{2} + 1) y^{^{'}} + 4 x (y - 1) = 0, y (0) = 5

\int x 16 - 25 x^{2} d x

\int_{0}^{\frac{π}{2}} 3 cos (x) d x

\int \frac{1}{( 4 x ^{2} - 25 ) ^{\frac{3}{2}}} d x

\int - 4 x^{2} - 12 x - 5 d x