+0  
 
0
671
3
avatar
Was ist die Ableitung von x*(x+1)^-1
 05.07.2015

Beste Antwort 

 #3
avatar+25739 
+8

Was ist die Ableitung von x*(x+1)^-1

 

1. Methode:

$$f(x) = x\cdot(x+1)^{-1}$$

$$\boxed{
~~f(x)=u\cdot v \qquad f'(x) = u\cdot v' + u'\cdot v ~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& (x+1)^{-1} \\
u' &=& 1 \quad & \quad v' &=& (-1)(x+1)^{-2}\cdot 1= -(x+1)^{-2}\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& x \cdot [-1(x+1)^{-2}] + 1\cdot (x+1)^{-1} \\
f'(x) &=& -x \cdot (x+1)^{-2} + (x+1)^{-1} \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \left(\dfrac{x+1}{x+1} \right) \\ \\
f'(x) &=& \dfrac{ -x +x+1} { (x+1)^2 } \\ \\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

 

2. Methode:

$$f(x) = \dfrac{ x } { x+1 }$$

$$\boxed{
~~f(x)=\dfrac{ u } { v } \qquad
f'(x) = \dfrac{ u } { v } \left(\dfrac{ u' } { u }-\dfrac{ v' } { v }
\right)~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& x+1 \\
u' &=& 1 \quad & \quad v' &=& 1\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1 } { x }-\dfrac{ 1 } {x+1} \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ x+1-x } { x(x+1) } \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1} { x(x+1) } \right)\\\\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

 

 06.07.2015
 #1
avatar+14537 
+3

$${f}{\left({\mathtt{x}}\right)} = {\frac{{\mathtt{x}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

 

f'(x) =   $${\frac{{\mathtt{1}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

 

Gruß radix !
 05.07.2015
 #2
avatar+12511 
+3

Omi67 05.07.2015
 #3
avatar+25739 
+8
Beste Antwort

Was ist die Ableitung von x*(x+1)^-1

 

1. Methode:

$$f(x) = x\cdot(x+1)^{-1}$$

$$\boxed{
~~f(x)=u\cdot v \qquad f'(x) = u\cdot v' + u'\cdot v ~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& (x+1)^{-1} \\
u' &=& 1 \quad & \quad v' &=& (-1)(x+1)^{-2}\cdot 1= -(x+1)^{-2}\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& x \cdot [-1(x+1)^{-2}] + 1\cdot (x+1)^{-1} \\
f'(x) &=& -x \cdot (x+1)^{-2} + (x+1)^{-1} \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \\ \\
f'(x) &=& \dfrac{ -x } { (x+1)^2 } + \dfrac{ 1 } { x+1 } \left(\dfrac{x+1}{x+1} \right) \\ \\
f'(x) &=& \dfrac{ -x +x+1} { (x+1)^2 } \\ \\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

 

2. Methode:

$$f(x) = \dfrac{ x } { x+1 }$$

$$\boxed{
~~f(x)=\dfrac{ u } { v } \qquad
f'(x) = \dfrac{ u } { v } \left(\dfrac{ u' } { u }-\dfrac{ v' } { v }
\right)~~
}$$

$$\begin{array}{rclrcl}
u &=& x \quad & \quad v &=& x+1 \\
u' &=& 1 \quad & \quad v' &=& 1\\\\
\end{array}\\\\
\begin{array}{rclrcl}
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1 } { x }-\dfrac{ 1 } {x+1} \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ x+1-x } { x(x+1) } \right)\\\\
f'(x) &=& \dfrac{ x } { x+1 } \left(\dfrac{ 1} { x(x+1) } \right)\\\\
\mathbf{f'(x)} &\mathbf{=}& \mathbf{\dfrac{1} { (x+1)^2 }} \\ \\
\end{array}$$

 

heureka 06.07.2015

42 Benutzer online