Man zeige mittels Differenzieren, dass
arcsin(x)=arctan(x1−x2){\displaystyle arcsin(x)=arctan\left({\frac {x}{\sqrt {1-x^{2}}}}\right)} für |x|<1{\displaystyle |x|<1}
ddxarctan(x)=11+x2{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arctan(x)={\frac {1}{1+x^{2}}}}
ddxarcsin(x)=11−x2{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}
ddxarctan(x1−x2)=ddxarctan(x⋅(1−x2)−12){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}arctan\left({\frac {x}{\sqrt {1-x^{2}}}}\right)={\frac {\mathrm {d} }{\mathrm {d} x}}arctan\left(x\cdot (1-x^{2})^{-{\frac {1}{2}}}\right)}
=11+(x1−x2)2⋅(1⋅(1−x2)−12+x⋅−12⋅(1−x2)−32⋅−2x){\displaystyle ={\frac {1}{1+({\frac {x}{\sqrt {1-x^{2}}}})^{2}}}\cdot \left(1\cdot (1-x^{2})^{-{\frac {1}{2}}}+x\cdot -{\frac {1}{2}}\cdot (1-x^{2})^{-{\frac {3}{2}}}\cdot -2x\right)}
=11+x21−x2⋅((1−x2)−12+x2⋅(1−x2)−32){\displaystyle ={\frac {1}{1+{\frac {x^{2}}{1-x^{2}}}}}\cdot \left((1-x^{2})^{-{\frac {1}{2}}}+x^{2}\cdot (1-x^{2})^{-{\frac {3}{2}}}\right)}
=11−x2+x21−x2⋅(......)=111−x2⋅(......)=1−x2⋅(......){\displaystyle ={\frac {1}{\frac {1-x^{2}+x^{2}}{1-x^{2}}}}\cdot \left(\;.\,.\,.\,.\,.\,.\;\right)={\frac {1}{\frac {1}{1-x^{2}}}}\cdot \left(\;.\,.\,.\,.\,.\,.\;\right)=1-x^{2}\cdot \left(\;.\,.\,.\,.\,.\,.\;\right)}
=(1−x2)⋅(1−x2)−12+x2⋅((1−x2)⋅(1−x2)−32){\displaystyle =(1-x^{2})\cdot (1-x^{2})^{-{\frac {1}{2}}}\;+\;x^{2}\cdot ((1-x^{2})\cdot (1-x^{2})^{-{\frac {3}{2}}})}
=(1−x2)12+x2⋅(1−x2)−12=1−x2+x21−x2=1−x2+x21−x2=11−x2{\displaystyle =(1-x^{2})^{\frac {1}{2}}+x^{2}\cdot (1-x^{2})^{-{\frac {1}{2}}}={\sqrt {1-x^{2}}}+{\frac {x^{2}}{\sqrt {1-x^{2}}}}={\frac {1-x^{2}+x^{2}}{\sqrt {1-x^{2}}}}={\frac {1}{\sqrt {1-x^{2}}}}}
→ddxarcsin(x)=ddxarctan(x1−x2)+C{\displaystyle \rightarrow \;{\frac {\mathrm {d} }{\mathrm {d} x}}\arcsin(x)={\frac {\mathrm {d} }{\mathrm {d} x}}arctan\left({\frac {x}{\sqrt {1-x^{2}}}}\right)+C}
q.e.d.
Manül 19:08, 27. Jun 2008 (CEST)
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