Lösung 1.2:3b

Aus Online Mathematik Brückenkurs 2

Version vom 10:05, 11. Mär. 2009

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

x−1x+1

and differentiating with the chain rule gives

ddxx−1x+1=12x−1x+1x−1x+1.

We establish the inner derivative by using the quotient rule,

ddxx−1x+1=12x−1x+1(x−1)2(x+1)(x−1)−(x+1)(x−1)=12x−1x+1(x−1)21(x−1)−(x+1)1=12x−1x+1−2(x−1)2=−x+1x−11(x−1)2=−1(x−1)32x+1

where we have used the simplification

x−1(x−1)2=(x−1)2(x−1)12=(x−1)12−2=(x−1)−32=1(x−1)32.