Ableitung für Kurvendiskussion

oh ja, bitte um step by step Lösung...bin schon gespannt....

[tex]
y=\frac {u(x)} {v(x)}
y'=\frac {u' v-v'u} {v^2}

y=\frac {x^3} {x^2-3}

u=x^3
u'=3x^2

v=x^2-3
v'=2x

y'=\frac {3x^2(x^2-3) - 2x*x^3} {(x^2-3)^2} = \frac {3x^4-9x^2-2x^4}{(x^2-3)^2} = \frac {x^4-9x^2} {(x^2-3)^2}

[/tex]

q.e.d.

zur zweiten Ableitung:
[tex]
y'= \frac {x^4-9x^2} {(x^2-3)^2}
u= x^4-9x^2
u'=4x^3-18x
v=(x^2-3)^2
v'=2(x^2-3)2x=4x(x^2-3)

y''=\frac {(4x^3-18x)(x^2-3)^2 - 4x(x^2-3)(x^4-9x^2)} {(x^2-3)^4}
=\frac {x^2-3}{x^2-3} * \frac {(4x^3-18x)(x^2-3) - 4x(x^4-9x^2)} { (x^2-3)^3}
= \frac {(4x^3-18x)(x^2-3) - 4x(x^4-9x^2)} { (x^2-3)^3}
=\frac {4x^5-12x^3-18x^3+54x-4x^4+36x^3} { (x^2-3)^3}
=\frac {6x^3+54x}{(x^2-3)^3} [/tex]
ὅπερ ἔδει δεῖξαι

und weiter zur dritten Ableitung:
[tex]
y''= \frac {6x^3+54x}{(x^2-3)^3}
u= 6x^3+54x
u'=18x^2+54
v=(x^2-3)^3
v'=3(x^2-3)^2*2x=6x(x^2-3)^2

y'''=\frac{(18x^2+54)(x^2-3)^3-6x(x^2-3)^2(6x^3+54x) }{(x^2-3)^6}
=\frac {(x^2-3)^2}{(x^2-3)^2} * \frac {(18x^2+54)(x^2-3)-6x(6x^3+54x)}{(x^2-3)^4}
= \frac {(18x^2+54)(x^2-3)-6x(6x^3+54x)}{(x^2-3)^4}
=\frac {-18x^4+324x^2-162} {(x^2-3)^4}


[/tex]

ὅπερ ἔδει ποιῆσαι