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Lösung 1.2:2f
Aus Online Mathematik Brückenkurs 2
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Version vom 10:05, 11. Mär. 2009
The entire expression is made up of several levels,
 1−x |
and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",
 |
and differentiate this using the chain rule,
 1−x=−sin1−x  1−x  . |
In the next differentiation, we have "the square root of something",
1−x=12 1−x(1−x) |
where we have used the differentiation rule,
| x=12x. |
for the outer derivative.
The whole differentiation in one go becomes
| 1−x=−sin1−xddx1−x=−sin1−x121−xddx(1−x)=−sin1−x121−x(−1)=21−xsin1−x. |
