(\partial)/(\partial w)(w/(h(w)^2))

Lösung

\frac{\partial}{\partial w} (\frac{w}{h ( w ) ( w ) ^{2}})

\frac{- h ( w ) - w \frac{\partial}{\partial w} ( h ( w ) )}{w ^{2} h ( w ) ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial w} (\frac{w}{h ( w ) ( w ) ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{\partial}{\partial w} ( w ) h ( w ) w ^{2} - \frac{\partial}{\partial w} ( h ( w ) w ^{2} ) w}{( h ( w ) w ^{2} ) ^{2}}

\frac{\partial}{\partial w} (w) = 1

\frac{\partial}{\partial w} (h (w) w^{2}) = \frac{\partial}{\partial w} (h (w)) w^{2} + 2 w h (w)

= \frac{1 \cdot h ( w ) w ^{2} - ( \frac{\partial}{\partial w} ( h ( w ) ) w ^{2} + 2 w h ( w ) ) w}{( h ( w ) w ^{2} ) ^{2}}

Vereinfache

\frac{1 \cdot h ( w ) w ^{2} - ( \frac{\partial}{\partial w} ( h ( w ) ) w ^{2} + 2 w h ( w ) ) w}{( h ( w ) w ^{2} ) ^{2}} : \frac{- h ( w ) - w \frac{\partial}{\partial w} ( h ( w ) )}{w ^{2} h ( w ) ^{2}}

= \frac{- h ( w ) - w \frac{\partial}{\partial w} ( h ( w ) )}{w ^{2} h ( w ) ^{2}}

\int_{0}^{2} (x) d x

\int e^{2 θ} + e^{2 θ} d θ

y^{- 2} \frac{d y}{d x} + y^{- 1} = 2 x

\frac{\partial}{\partial x} (e^{3 x} tan (y))

y^{^{''}} + 6 y^{^{'}} - 16 y = 0