derivative f(t)=(t^2)/(sqrt(t^3+1))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d t} (t^{2}) = 2 t

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

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

Vereinfache

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

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

d er i v a t i v e \frac{y ^{\frac{3}{2}}}{3} - y^{\frac{1}{2}}

\frac{d}{d x} (e^{2 x} - x)

(\frac{1}{x ^{4}})^{^{'}}

\frac{\partial}{\partial x} (4 - x^{4} + 2 x^{2} - y^{2})

\frac{\partial}{\partial x} (4 x - 2 y^{2} + z^{2})