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Lösung 1.2:3c

Aus Online Mathematik Brückenkurs 2

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Version vom 10:06, 11. Mär. 2009

We can write the expression as

1x1x2=x1x21,

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

ddxx1x21=1x1x22x1x2=1x1x22x1x2=1x2(1x2)x1x2.

The next step is to differentiate the product x1x2  using the product rule,

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

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

=1x2(1x2)1x2+x121x2(1x2)=1x2(1x2)1x2+x121x2(2x)=1x2(1x2)1x2x21x2.

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

=1x2(1x2)1x21x22x2=1x2(1x2)1x21x2x2=12x2x2(1x2)32.


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