TU Wien:Mathematik 1 UE (diverse)/Übungen SS08/Beispiel 76

Man zeige mittels Differenzieren, dass

$arcsin(x)=arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$ für $|x|<1$

Benötigt

$\frac{\mathrm{d}}{\mathrm{d}x}\arctan (x)=\frac{1}{1+x^2}$

Lösungsvorschlag

$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}$

$\frac{\mathrm{d}}{\mathrm{d}x}arctan\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{\mathrm{d}}{\mathrm{d}x}arctan(x\cdot (1-x^2{)}^{-\frac{1}{2}})$

$=\frac{1}{1+(\frac{x}{\sqrt{1-x^2}}{)}^2}\cdot (1\cdot (1-x^2{)}^{-\frac{1}{2}}+x\cdot -\frac{1}{2}\cdot (1-x^2{)}^{-\frac{3}{2}}\cdot -2x)$

$=\frac{1}{1+\frac{x^2}{1-x^2}}\cdot ((1-x^2{)}^{-\frac{1}{2}}+x^2\cdot (1-x^2{)}^{-\frac{3}{2}})$

$=\frac{1}{\frac{1-x^2+x^2}{1-x^2}}\cdot (......)=\frac{1}{\frac{1}{1-x^2}}\cdot (......)=1-x^2\cdot (......)$

$=(1-x^2)\cdot (1-x^2{)}^{-\frac{1}{2}}+x^2\cdot ((1-x^2)\cdot (1-x^2{)}^{-\frac{3}{2}})$

$=(1-x^2{)}^{\frac{1}{2}}+x^2\cdot (1-x^2{)}^{-\frac{1}{2}}=\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}=\frac{1-x^2+x^2}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$

$\to \frac{\mathrm{d}}{\mathrm{d}x}\arcsin (x)=\frac{\mathrm{d}}{\mathrm{d}x}arctan\left(\frac{x}{\sqrt{1-x^2}}\right)+C$

q.e.d.

Manül 19:08, 27. Jun 2008 (CEST)