Seite 17 Nr. 5c) g ( x ) = x + 1 x − 1 {\displaystyle g(x)={\frac {\sqrt {x+1}}{\sqrt {x-1}}}} u ( x ) = x + 1 ↦ u ′ ( x ) = 1 2 ∗ x + 1 {\displaystyle u(x)={\sqrt {x+1}}\mapsto u\,'(x)={\frac {1}{2*{\sqrt {x+1}}}}} v ( x ) = x − 1 ↦ v ′ ( x ) = 1 2 ∗ x − 1 {\displaystyle v(x)={\sqrt {x-1}}\mapsto v\,'(x)={\frac {1}{2*{\sqrt {x-1}}}}} g ′ ( x ) = ( 1 2 ∗ x + 1 ) ∗ ( x − 1 ) − ( x + 1 ) ∗ ( 1 2 ∗ x − 1 ) x − 1 {\displaystyle g\,'(x)={\frac {({\frac {1}{2*{\sqrt {x+1}}}})*({\sqrt {x-1}})-({\sqrt {x+1}})*({\frac {1}{2*{\sqrt {x-1}}}})}{x-1}}} nächster schritt: = x − 1 2 x + 1 − x + 1 2 x − 1 x − 1 {\displaystyle ={\frac {{\frac {\sqrt {x-1}}{2{\sqrt {x+1}}}}-{\frac {\sqrt {x+1}}{2{\sqrt {x-1}}}}}{x-1}}} (ausmultipliziert) = 2 x − 1 ∗ x − 1 − 2 ∗ x + 1 ∗ x + 1 2 ∗ x + 1 ∗ ( 2 x − 1 ) x − 1 {\displaystyle ={\frac {\frac {2{\sqrt {x-1}}*{\sqrt {x-1}}-2*{\sqrt {x+1}}*{\sqrt {x+1}}}{2*{\sqrt {x+1}}*(2{\sqrt {x-1}})}}{x-1}}} (erweitert) = 2 ( x − 1 ) − 2 ( x + 1 ) 4 x + 1 ∗ x − 1 ∗ ( x − 1 ) {\displaystyle ={\frac {2(x-1)-2(x+1)}{4{\sqrt {x+1}}*{\sqrt {x-1}}*(x-1)}}} (vereinfacht)
So nu geändert -.-
