derivative (2.25r)/((r+0.5)^2)

Lösung

ableitung von

\frac{2.25 r}{( r + 0.5 ) ^{2}}

\frac{2.25 ( 0.5 - r )}{( r + 0.5 ) ^{3}}

Schritte zur Lösung

\frac{d}{d r} (\frac{2.25 r}{( r + 0.5 ) ^{2}})

Entferne die Konstante:

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

= 2.25 \frac{d}{d r} (\frac{r}{( r + 0.5 ) ^{2}})

Wende die Quotientenregel an:

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

= 2.25 \cdot \frac{\frac{d r}{d r} ( r + 0.5 ) ^{2} - \frac{d}{d r} ( ( r + 0.5 ) ^{2} ) r}{( ( r + 0.5 ) ^{2} ) ^{2}}

\frac{d r}{d r} = 1

\frac{d}{d r} ((r + 0.5)^{2}) = 2 (r + 0.5)

= 2.25 \cdot \frac{1 \cdot ( r + 0.5 ) ^{2} - 2 ( r + 0.5 ) r}{( ( r + 0.5 ) ^{2} ) ^{2}}

Vereinfache

2.25 \cdot \frac{1 \cdot ( r + 0.5 ) ^{2} - 2 ( r + 0.5 ) r}{( ( r + 0.5 ) ^{2} ) ^{2}} : \frac{2.25 ( 0.5 - r )}{( r + 0.5 ) ^{3}}

= \frac{2.25 ( 0.5 - r )}{( r + 0.5 ) ^{3}}

\int (x^{2} + x + 1) d x

\int cos (θ) sin^{2} (θ) d θ

\frac{\partial}{\partial y} (x^{2} cos (e^{x y}))

y^{^{''}} + 6 y^{^{'}} + 9 y = - x e^{4 x}

\int \frac{8 x}{x ^{2} + x + 1} d x