Lösung 1.2:4a

Aus Online Mathematik Brückenkurs 2

Version vom 10:07, 11. Mär. 2009

We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives

ddxx1−x2=1−x22(x)1−x2−x1−x2=1−x211−x2−x1−x2.

We determine the derivative 1−x2  by using the chain rule

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

We simplify the result as far as possible, so as to make the second differentiation easier,

=1−x21−x2+x21−x2=1−x21−x21−x22+x21−x2=1−x21−x21−x2+x2=1(1−x2)32.

The second derivative is

d2dx2x1−x2=ddx1(1−x2)32=ddx1−x2−32=−231−x2−32−11−x2=−231−x2−52(−2x)=3x1−x2−52=3x1−x252.