derivative 6+(2t)/(1+t^3)-e^{-0.02t}

Lösung

ableitung von

6 + \frac{2 t}{1 + t ^{3}} - e^{- 0.02 t}

0.02 e^{- 0.02 t} + \frac{2 ( - 2 t ^{3} + 1 )}{( t ^{3} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (6 + \frac{2 t}{1 + t ^{3}} - e^{- 0.02 t})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (6) + \frac{d}{d t} (\frac{2 t}{1 + t ^{3}}) - \frac{d}{d t} (e^{- 0.02 t})

\frac{d}{d t} (6) = 0

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

\frac{d}{d t} (e^{- 0.02 t}) = - 0.02 e^{- 0.02 t}

= 0 + \frac{2 ( 1 - 2 t ^{3} )}{( 1 + t ^{3} ) ^{2}} - (- 0.02 e^{- 0.02 t})

Vereinfache

= 0.02 e^{- 0.02 t} + \frac{2 ( - 2 t ^{3} + 1 )}{( t ^{3} + 1 ) ^{2}}

\frac{d}{d s} (e^{- s})

\int 5 cos (\frac{x}{2}) d x

d er i v a t i v e x^{2} 16 - x^{2}

in v erse l a pl a ce \frac{1}{s ^{2} - 2 s}

d er i v a t i v e - \frac{8}{x ^{2}}