derivative f(s)=sqrt((s^2+1)/(s^2+4))

Lösung

ableitung von

f (s) = \frac{s ^{2} + 1}{s ^{2} + 4}

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

Schritte zur Lösung

\frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

Wende die Kettenregel an:

\frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

= \frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

\frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4}) = \frac{6 s}{( s ^{2} + 4 ) ^{2}}

= \frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \cdot \frac{6 s}{( s ^{2} + 4 ) ^{2}}

Vereinfache

\frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \cdot \frac{6 s}{( s ^{2} + 4 ) ^{2}} : \frac{3 s}{s ^{2} + 1 ( s ^{2} + 4 ) ^{\frac{3}{2}}}

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

\frac{\partial}{\partial x} ((x^{2} + 5 x y - 2 y)^{8})

l a pl a ce t r an s f or m e^{- 5 t} sin (7 t)

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