Hallo Azion,
f(x) = √(2x+1) * √(x-1) (?)
Produktregel: [ u*v ] ' = u ' • v + u • v '
[√x] ' = 1 / (2·√x)
Kettenregel hier: [ √u ] ' = 1 / (2·√u) • u' = u' / (2·√u)
f '(x) = \(\frac{1}{2·√(2x+1)}\) · 2 · √(x-1) + √(2x+1) ·\(\frac{1}{2·√(x-1)}\)
= \(\frac{√(x-1)}{√(2x+1)}\) + \(\frac{√(2x+1)}{2·√(x-1)}\)
= \(\frac{2·(x-1) + 2x+1}{2·√(2x+1)·√(x-1) ] }\) = \(\frac{4x-1}{2·√(2x+1)·√(x-1)}\)
Gruß Wolfgang
