Lagrange Interpolation and derivative

Question

$f:[a,b] \to R$ is a continuous differentiable function, and all $x_i$ are different and $q_i(x) = \prod (x-x_j)^2$ with $i \not= j$

Show that $$g_i(x)= \frac{q_i(x)}{q_i(x_i)}\left(\left(1-\frac{(x-x_i)q_i'(x_i)}{q_i(x_i)}\right)f(x_i)+(x-x_i)\cdot f'(x_i)\right)$$

interpolates $f$ and $f'$ in $x_i$.

For $f$ I just have to calculate $g_i(x_i)$, but how can I show that for $f'$?

Answer 1

Trivially by this formula we can see that $g_i(x_i)=f(x_i)$. To solve the second part just differentiate the function $g_i$ with respect to $x$ and then put $x_i$ in a place of $x$ (compute $g_i'(x_i)$).

Let $g$ be the interpolant of $f$ at the double node $x_0$. This means that $g(x_0)=f(x_0)$ and $g'(x_0)=f'(x_0)$. That is why you should differentiate $g_i$.