We have $f(x)=log(1+x)$, and $x_{}=0, x_{1}=1, x_{2}=t$ with obviously $t>1$ and $x>-1$
a) find the interpolation polynomial using Newton's method.
I found : $$p_{2}(x)=x \cdot log(2) + x \cdot (x-1) \cdot \frac{\frac{log(1+t)-log(2)}{t-1}-log(2)}{t}$$
b) To what corresponds $\delta f[x_{1},x_{2}]$ for $t \rightarrow 1$.
I found : $$\lim_{t \rightarrow 1} \frac{log(1+t)-log(2)}{t-1} = \lim_{t \rightarrow 0} \frac{log(2+t)-log(2)}{t} = f'(x_{1})$$
c) Write $h_{2}(x)$, the interpolation polynomial when we have $t \rightarrow 1$.
I found : $$h_{2}(x)=x \cdot log(2) + x \cdot (x-1) \cdot (\frac{1}{2}-log(2))$$
$$\ \ \ \ \ \ \ \ \ \ \ \ =x \cdot f(x_{1}) + (x^{2}-x) \cdot (f'(x_{1})-f(x_{1}))$$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f(x_{1})\cdot(-x^{2} + 2 \cdot x) + f'(x_{1})\cdot(x^{2}-x)$$
And now they ask me :
What is the link between the polynomial interpolation and the derivative of $f(x)$ in $x_{1}$
And here I just have no idea.. Any help would be great thanks