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

Lösung

ableitung von

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

- \frac{2 x ( - x ^{2} - 12 )}{( x ^{2} - 4 ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{( - 2 x ) ( x ^{2} - 4 ) ^{2} - 4 x ( x ^{2} - 4 ) ( - x ^{2} - 4 )}{( ( x ^{2} - 4 ) ^{2} ) ^{2}} : - \frac{2 x ( - x ^{2} - 12 )}{( x ^{2} - 4 ) ^{3}}

= - \frac{2 x ( - x ^{2} - 12 )}{( x ^{2} - 4 ) ^{3}}

n = 0 \sum \infty \frac{1}{( 1 + n ^{2} )}

n = 0 \sum \infty \frac{1}{7 ^{n}}

\frac{d y}{d x} = \frac{( x + sec ^{2} ( x ) )}{y}

\int \frac{x}{x ^{4} + 16} d x

a re a y = 3 x^{2} - 12, y = 0