derivative h(t)=((t^3)/(t^8+9))^2

Lösung

ableitung von

h (t) = (\frac{t ^{3}}{t ^{8} + 9})^{2}

\frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

Schritte zur Lösung

\frac{d}{d t} ((\frac{t ^{3}}{t ^{8} + 9})^{2})

Wende die Kettenregel an:

2 \cdot \frac{t ^{3}}{t ^{8} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9})

= 2 \cdot \frac{t ^{3}}{t ^{8} + 9} \frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9})

\frac{d}{d t} (\frac{t ^{3}}{t ^{8} + 9}) = \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}}

= 2 \cdot \frac{t ^{3}}{t ^{8} + 9} \cdot \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}}

Vereinfache

2 \cdot \frac{t ^{3}}{t ^{8} + 9} \cdot \frac{- 5 t ^{10} + 27 t ^{2}}{( t ^{8} + 9 ) ^{2}} : \frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

= \frac{2 t ^{3} ( - 5 t ^{10} + 27 t ^{2} )}{( t ^{8} + 9 ) ^{3}}

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