derivative (4-t)(2+t^2)^{-1}

Lösung

ableitung von

(4 - t) (2 + t^{2})^{- 1}

\frac{t ^{2} - 8 t - 2}{( t ^{2} + 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} ((4 - t) (2 + t^{2})^{- 1})

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 4 - t, g = (2 + t^{2})^{- 1}

= \frac{d}{d t} (4 - t) (2 + t^{2})^{- 1} + \frac{d}{d t} ((2 + t^{2})^{- 1}) (4 - t)

\frac{d}{d t} (4 - t) = - 1

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

= (- 1) (2 + t^{2})^{- 1} + (- \frac{2 t}{( 2 + t ^{2} ) ^{2}}) (4 - t)

Vereinfache

(- 1) (2 + t^{2})^{- 1} + (- \frac{2 t}{( 2 + t ^{2} ) ^{2}}) (4 - t) : \frac{t ^{2} - 8 t - 2}{( t ^{2} + 2 ) ^{2}}

= \frac{t ^{2} - 8 t - 2}{( t ^{2} + 2 ) ^{2}}

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