derivative g(t)=(9t+5)^2(5t-6)^{-3}

Lösung

ableitung von

g (t) = (9 t + 5)^{2} (5 t - 6)^{- 3}

\frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

Schritte zur Lösung

\frac{d}{d t} ((9 t + 5)^{2} (5 t - 6)^{- 3})

Wende die Produktregel an:

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

f = (9 t + 5)^{2}, g = (5 t - 6)^{- 3}

= \frac{d}{d t} ((9 t + 5)^{2}) (5 t - 6)^{- 3} + \frac{d}{d t} ((5 t - 6)^{- 3}) (9 t + 5)^{2}

\frac{d}{d t} ((9 t + 5)^{2}) = 18 (9 t + 5)

\frac{d}{d t} ((5 t - 6)^{- 3}) = - \frac{15}{( 5 t - 6 ) ^{4}}

= 18 (9 t + 5) (5 t - 6)^{- 3} + (- \frac{15}{( 5 t - 6 ) ^{4}}) (9 t + 5)^{2}

Vereinfache

18 (9 t + 5) (5 t - 6)^{- 3} + (- \frac{15}{( 5 t - 6 ) ^{4}}) (9 t + 5)^{2} : \frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

= \frac{- 405 t ^{2} - 1872 t - 915}{( 5 t - 6 ) ^{4}}

(3 x + 2) + (3 y - 3) y^{^{'}} = 0

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