Suppose, we have the data points $f_a,f_b$ as well as $f_a',f_b'$. We have a cubic spline $P(x)=c(x-a)^3+d(x-a)^2+e(x-a)+f$ in $[a,b]$ where:$$P(a)=f_a, \, P(b)=f_b,\, P'(a)=f_a' \, \text{ und } P'(b)=f_b'$$
I want to show that this setting will give a unique polynomial.
My attempt
We have $P'(x)=3c(x-a)^2+2d(x-a)+e$ and thus:$$ \begin{cases}P(a)=f=f_a \\ P(b)=c(b-a)^3+d(b-a)^2+e(b-a)+f_a=f_b \\ P'(a)=e=f_a' \\ P'(b)=3c(b-a)^2+2d(b-a)+f_a'=f_b'\end{cases}$$
How do I know that this will yield a unique $P$? How can I solve this system of equations?
Important note We have $b>a$.
A) First step : identifying your expression
$$P(x)=c(x-a)^3+d(x-a)^2+e(x-a)+f$$
with (exact) Taylor expansion of $P$ in the vicinity of $a$:
$$P(x)=P(a)+P'(a)(x-a)+\frac12 P''(a)(x-a)^2+ \frac16 P'''(a)(x-a)^3$$
already gives two coefficients :
B) Second step : Therefore, it remains to find the two unknowns $c,d$ in expression :
$$P(x)=c(x-a)^3+d(x-a)^2+f'_a(x-a)+f_a$$
taking account of
a) $P(b)=f_b \ \iff \ c(b-a)^3+d(b-a)^2+f'_a(b-a)+f_a=f_b$
b) $P'(b)=f'_b \ \iff \ 3c(b-a)^2+2d(b-a)+f'_a=f'_b$
Therefore, we have a linear system of 2 equations with 2 unknowns $c,d$:
$$\begin{cases}c(b-a)^3+d(b-a)^2&=&f_b-f_a-f'_a(b-a)\\3c(b-a)^2+2d(b-a)&=&f'_b-f'_a\end{cases}$$
Said otherwise :
$$\begin{cases}c(b-a)+d&=&\frac{f_b-f_a-f'_a(b-a)}{(b-a)^2}\\3c(b-a)+2d&=&\frac{f'_b-f'_a}{(b-a)}\end{cases}$$
Can you take it from here ?