d/(da)((pi(9a^2+6a+1))/(36a))

Lösung

\frac{d}{d a} (\frac{π ( 9 a ^{2} + 6 a + 1 )}{36 a})

\frac{π ( 9 a ^{2} - 1 )}{36 a ^{2}}

Schritte zur Lösung

\frac{d}{d a} (\frac{π ( 9 a ^{2} + 6 a + 1 )}{36 a})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{π}{36} \frac{d}{d a} (\frac{9 a ^{2} + 6 a + 1}{a})

Wende die Quotientenregel an:

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

= \frac{π}{36} \cdot \frac{\frac{d}{d a} ( 9 a ^{2} + 6 a + 1 ) a - \frac{d a}{d a} ( 9 a ^{2} + 6 a + 1 )}{a ^{2}}

\frac{d}{d a} (9 a^{2} + 6 a + 1) = 18 a + 6

\frac{d a}{d a} = 1

= \frac{π}{36} \cdot \frac{( 18 a + 6 ) a - 1 \cdot ( 9 a ^{2} + 6 a + 1 )}{a ^{2}}

Vereinfache

\frac{π}{36} \cdot \frac{( 18 a + 6 ) a - 1 \cdot ( 9 a ^{2} + 6 a + 1 )}{a ^{2}} : \frac{π ( 9 a ^{2} - 1 )}{36 a ^{2}}

= \frac{π ( 9 a ^{2} - 1 )}{36 a ^{2}}

n = 0 \sum \infty 1.0 5^{n}

\frac{d}{d x} (π sin (π x))

in v erse l a pl a ce \frac{0.26 s}{( s + 1 )}

a re a 16 - x, x + 4, 0, - 9

d er i v a t i v e y = (sin (x)) (cos (x))