f(2)=√3·2–5=1
f`(x)=(3x–5)`/(2·√3x–5)
f`(x)=3/(2·√3x–5
f`(2)=3/2
f``(x)=(3/2)·((3x–5)–1/2)`=(3/2)·(–1/2)·(3x–5)–3/2·(3x–5)`
f``(x)=(–9/4)·(3x–5)–3/2
f``(2)=–9/4
f```(x)=(–9/4)·(–3/2)·(3x–5)–5/2·(3x–5)`
f```(x)=(81/8)(3x–5)–5/2
f```(2)=81/8
f(x) ~ f(xo) + (f`(xo)/1!)·(x–xo)2 + (f``(xo)/2!)·(x–xo)2 +
+ (f```(xo)/3!)·(x–xo)3 +...
√3x–5 ~ 1+(3/2)·(x–2) – (9/8)·(x–2)2 + (81/48)·(x–2)3