\(\begin{equation}\begin{split}
f''(x)&=\dfrac{(-8x^3+6x)\cdot ((1-2x^2)^2) - (-2x^4+3x^2) \cdot (-8 x (1 - 2 x^2))}{(1-2x^2)^4}
\\&= \dfrac{(-8x^3+6x)\cdot ((1-2x^2)^2) +8 x (1 - 2 x^2) (3 x^2 - 2 x^4)}{(1-2x^2)^4}
\\&= \dfrac{(-8x^3+6x)\cdot ((1-2x^2)^2)}{(1-2x^2)^4} + \dfrac{8 x (1 - 2 x^2) (3 x^2 - 2 x^4)}{(1-2x^2)^4}
\\&= \dfrac{(-8x^3+6x)\cdot (1-2x^2)}{(1-2x^2)^3} + \dfrac{8 x (3 x^2 - 2 x^4)}{(1-2x^2)^3}
\\&= \dfrac{(-8x^3+6x)\cdot (1-2x^2) + 8 x (3 x^2 - 2 x^4)}{(1-2x^2)^3}
\\&= \dfrac{[16 x^5 - 20 x^3 + 6 x] + [- 16 x^5+24 x^3]}{(1-2x^2)^3}
\\&= \dfrac{4x^3+6x}{(1-2x^2)^3}
\end{split}\end{equation}\)
Sonstiger Berufsstatus, Punkte: 16.5K
Grüße Manuel ─ manuel_h 01.08.2019 um 18:20