I have calculated the tangent for start point and end point of cubic bezier curve, but the calculated slope and coordinates of tangents are not continuous to the tangent lines (segment $P_0 P_1$ and $P_2 P_3$), as shown in this picture. $m_{start}$ and $m_{end}$ from calculated tangent (blue lines in picture) and tangents which are the subtraction of control points are different. Is there something wrong in my calculation?
The curve equation,
$ C(t) = (1-t)^3P_0 + 3t(1-t)^2P_1+3t^2(1-t)P_2 + t^3P_3 $
Control points are:
$$ P_0 = (14.89, 118.65) $$ $$ P_1 = (40.40, 130.86) $$ $$ P_2 = (65.91, 143.08) $$ $$ P_3 = (71.44, 124.94) $$
Derivative of curve, $$ C'(t) = -3(1-t)^2 P_0 + \left[3(1-t)^2 -6t(1-t)\right]P_1 + \left[6t(1-t)-3t^2\right]P_2 + 3t^2P_3 $$
Substituting the coordinate value of control points in C'(t), $$ x = -59.94t^2 + 76.53 $$ $$ y = -91.02t^2 - 0.06t + 36.66 $$
Tangent at $ t $: $$ T(t) = \frac {C'(t)} {\vert C'(t) \vert} = \frac {C'(t)_x, C'(t)_y} {\sqrt {\left[C'(t)_x\right]^2 + \left[C'(t)_y\right]^2}} $$
Tangent at end point, $ t=1 $: $$ T(1)_x = 0.29 $$ $$ T(1)_y = -0.95 $$
Tangent at start point, $ t = 0 $: $$ T(0)_x = 0.9 $$ $$ T(0)_y = -0.43 $$
Slope equation: $$ m = \frac {y-y_0} {x-x_0} $$
Slope for end tangent, t = 1: $$ m_{end} = \frac {P_{3y}-T(1)_y}{P_{3x}-T(1)_x} = 1.77 $$
Slope for start tangent, t = 0: $$ m_{start} = \frac {P_{0y}-T(0)_y}{P_{0x}-T(0)_x} = 8.45 $$
slope for segment $P_0P_1$,
$$ m_{start} = \frac {P_{0y}-P_{1y}}{P_{0x}-P_{1x}} = 0.47 $$
slope for segment $P_2P_3$,
$$ m_{end} = \frac {P_{2y}-P_{3y}}{P_{2x}-P_{3x}} = -3.28 $$
Tangent at $t=0$ should be $(0.902, 0.4320)$.
From $T(0)$ and $T(1)$, we can compute the slope as $0.432/0.902=0.478$ at $t=0$ and $-0.95/0.29=-3.27$, which matches your computation of slopes from $P_0P_1$ and $P_2P_3$