d/(dt)((x)(2/(t^2+1)-1))

Lösung

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

\frac{- t ^{4} \frac{d x}{d t} - 4 t x + \frac{d x}{d t}}{( t ^{2} + 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = \frac{2}{t ^{2} + 1} - 1

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

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

= \frac{d x}{d t} (\frac{2}{t ^{2} + 1} - 1) + (- \frac{4 t}{( t ^{2} + 1 ) ^{2}}) x

Vereinfache

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

= \frac{- t ^{4} \frac{d x}{d t} - 4 t x + \frac{d x}{d t}}{( t ^{2} + 1 ) ^{2}}

e x p an d \frac{( 7 q ^{2} + 4 ) ^{2}}{49}

e x p an d \frac{5 ^{n - 1}}{( n + 1 ) ^{2} 4 ^{n}}

e x p an d 5 (x^{2} + y^{2}) e^{y^{2} - x^{2}}

e x p an d \frac{a ^{2} + 4 a + 4}{( 3 a + 3 b ) ( a + b )}

e x p an d \frac{3 x ^{2} + 1}{( x + 2 ) ^{3}}