derivative f(t)=(3t-1)^4(2t+1)^{-3}

Lösung

ableitung von

f (t) = (3 t - 1)^{4} (2 t + 1)^{- 3}

\frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

Schritte zur Lösung

\frac{d}{d t} ((3 t - 1)^{4} (2 t + 1)^{- 3})

Wende die Produktregel an:

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

f = (3 t - 1)^{4}, g = (2 t + 1)^{- 3}

= \frac{d}{d t} ((3 t - 1)^{4}) (2 t + 1)^{- 3} + \frac{d}{d t} ((2 t + 1)^{- 3}) (3 t - 1)^{4}

\frac{d}{d t} ((3 t - 1)^{4}) = 12 (3 t - 1)^{3}

\frac{d}{d t} ((2 t + 1)^{- 3}) = - \frac{6}{( 2 t + 1 ) ^{4}}

= 12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4}

Vereinfache

12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4} : \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

= \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

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