Lösung 1.2:3c

Aus Online Mathematik Brückenkurs 2

We can write the expression as

1x1−x2=x1−x2−1,

and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,

ddxx1−x2−1=−1x1−x2−2x1−x2=−1x1−x22x1−x2=−1x2(1−x2)x1−x2.

The next step is to differentiate the product x1−x2  using the product rule,

=−1x2(1−x2)(x)1−x2+x(1−x2)=−1x2(1−x2)11−x2+x(1−x2).

The expression 1−x2  is of the type "square root of something", so we use the chain rule to differentiate,

=−1x2(1−x2)1−x2+x121−x2(1−x2)=−1x2(1−x2)1−x2+x121−x2(−2x)=−1x2(1−x2)1−x2−x21−x2.

We write the expression on the right over a common denominator,

=−1x2(1−x2)1−x21−x22−x2=−1x2(1−x2)1−x21−x2−x2=−1−2x2x2(1−x2)32.

Note: When we make simplifications of the form 1−x22=1−x2 , we assume that both sides are well defined (i.e. in this case that x lies between -1 and 1).