derivative ((3t^2+4))/(4t+4)

Lösung

ableitung von

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

\frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d t} (3 t^{2} + 4) = 6 t

\frac{d}{d t} (4 t + 4) = 4

= \frac{6 t ( 4 t + 4 ) - 4 ( 3 t ^{2} + 4 )}{( 4 t + 4 ) ^{2}}

Vereinfache

\frac{6 t ( 4 t + 4 ) - 4 ( 3 t ^{2} + 4 )}{( 4 t + 4 ) ^{2}} : \frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

= \frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

\int \frac{csc ( θ ) cot ( θ )}{9} d θ

\frac{d}{d x} (x^{2} 4 x - 9)

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u

d er i v a t i v e (3 x^{2} - 10)^{- 13}

\int (- 3 (\frac{9 x ^{2}}{2}) - 4 x^{3}) d x