Lösung 1.2:3b
Aus Online Mathematik Brückenkurs 2
The outer function in the expression is "the square root of something",
| \displaystyle \sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } |
and differentiating with the chain rule gives
| \displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.} |
We establish the inner derivative by using the quotient rule,
| \displaystyle \begin{align}
\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}} &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt] &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt] &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt] &= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt] &= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,, \end{align} |
where we have used the simplification
| \displaystyle \frac{\sqrt{x-1}}{(x-1)^2}
= \frac{(x-1)^{1/2}}{(x-1)^2} = (x-1)^{1/2-2} = (x-1)^{-3/2} = \frac{1}{(x-1)^{3/2}}\,\textrm{.} |
