Příklad 2.6
with(plots) :
with (Student) :
with(Student[Calculusl]) :
fln(2-x3+4-x2-x) \ J'    V           x+1             )'
fl:=diff(f,x);
2                                                 /       3             2           i
_____6x + 8x — 1_______ ln^2x +4x —xj
(2x3+4x2-x) (x + 1)             (x+1)2
ß:=dijf(f,x$2);
_______12x + 8_____________(6x2 + 8x-l)2__________2 (óx2 + 8x-l)
(2x3+4x2-x) (x + 1)      (2x3+4x2-xf (x + 1)      (2x3+4x2-x) (x + l)2
21n(2x3+4x2-x)
(x+1)3
s;
w
N3
I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I
■d)   '    '    '    '
i
N>
—  INJ
— w
— J*.
I— Ul
DerivativePlotif, x = 0.9 ..5, thickness = 2 );
The Derivativeof
f(x)= ln(2*xA3+4*xA2-x)/(x+1)
on the Interval[.9, 5]
f(x)
1 st derivative
lO —I
^ = „
■sř-
	(D	0
	O	O
M—	CO	CO
(1)	>	>
O	!_	s_
*Ĺ	0	0
C	T3	T3
3		
CNI
A\]:=eval{f,x=\);
|ln(5)                                                                    (1)
fl[\]:=eval{fl,x=\);
H-^ln(5)                                                   (2)
f2[\]:=eval(fl,x=\);
-|+^ln(5)                                                             (3)
evfl//(/[l]);
0.8047189560                                                               (4)
evalf(fl[l]);
0.8976405220                                                               (5)
evalf(fi[l]);
-2.277640522                                                               (6)
showtangentif, x = 1, x = 0.5 ..5, co/or = [cyan, red], thickness = 2);
X