d/(dy)(y/(y+b/y))

Lösung

\frac{d}{d y} (\frac{y}{y + \frac{b}{y}})

\frac{2 b y}{( y ^{2} + b ) ^{2}}

Schritte zur Lösung

\frac{d}{d y} (\frac{y}{y + \frac{b}{y}})

Wende die Quotientenregel an:

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

= \frac{\frac{d y}{d y} ( y + \frac{b}{y} ) - \frac{d}{d y} ( y + \frac{b}{y} ) y}{( y + \frac{b}{y} ) ^{2}}

\frac{d y}{d y} = 1

\frac{d}{d y} (y + \frac{b}{y}) = 1 - \frac{b}{y ^{2}}

= \frac{1 \cdot ( y + \frac{b}{y} ) - ( 1 - \frac{b}{y ^{2}} ) y}{( y + \frac{b}{y} ) ^{2}}

Vereinfache

\frac{1 \cdot ( y + \frac{b}{y} ) - ( 1 - \frac{b}{y ^{2}} ) y}{( y + \frac{b}{y} ) ^{2}} : \frac{2 b y}{( y ^{2} + b ) ^{2}}

= \frac{2 b y}{( y ^{2} + b ) ^{2}}

n = 1 \sum \infty - \frac{1}{3 ( 3 n + 2 )}

\int \frac{1}{y ln ( y )} d y

\frac{\partial}{\partial v} (u + 2 v)

\int_{1}^{2} \frac{v ^{6} + 4 v ^{8}}{v ^{5}} d v

y^{^{''}} + 4 y^{^{'}} - 16 y = 0