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

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:

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

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

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

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

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 !