derivative ((t^6))/((t^3+1)^2)

Lösung

ableitung von

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

\frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d t} (t^{6}) = 6 t^{5}

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

= \frac{6 t ^{5} ( t ^{3} + 1 ) ^{2} - 6 t ^{2} ( t ^{3} + 1 ) t ^{6}}{( ( t ^{3} + 1 ) ^{2} ) ^{2}}

Vereinfache

\frac{6 t ^{5} ( t ^{3} + 1 ) ^{2} - 6 t ^{2} ( t ^{3} + 1 ) t ^{6}}{( ( t ^{3} + 1 ) ^{2} ) ^{2}} : \frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

= \frac{6 t ^{5}}{( t ^{3} + 1 ) ^{3}}

\int (1 - sin (θ))^{2} d θ

\int cos (- 1 (x)) d x

\int 2 x (x^{2} + 4)^{5} d x

d er i v a t i v e y = \frac{6 x}{e ^{x}}

\frac{d}{d x} ((x - 1)^{\frac{3}{2}})