Aus Online Mathematik Brückenkurs 2

The outer function in the expression is "the root of something",

\displaystyle \sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}

and differentiating with the chain rule gives

\displaystyle \frac{d}{dx}\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}}\centerdot \left( \frac{x+1}{x-1} \right)^{\prime }

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{\frac{x+1}{x-1}}}\centerdot \frac{\left( x+1 \right)^{\prime }\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot \left( x-1 \right)^{\prime }}{\left( x-1 \right)^{2}} \\ & =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{1\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot 1}{\left( x-1 \right)^{2}} \\ & =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\ & =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\ & =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\ \end{align}

where we have used the simplification

\displaystyle \frac{\sqrt{x-1}}{\left( x-1 \right)^{2}}=\frac{\left( x-1 \right)^{{1}/{2}\;}}{\left( x-1 \right)^{2}}=\left( x-1 \right)^{\frac{1}{2}-2}=\left( x-1 \right)^{-\frac{3}{2}}=\frac{1}{\left( x-1 \right)^{\frac{3}{2}}}