How to take the directional derivative of an integral over $u$ from $(x,y)$ to $(1,1)$ Please consider the following function
$$
v(x,y) = \int_{(x,y)}^{(1, 1)}u\,dS
$$
$v(x,y)$ is the integral along the straight line from $(x,y)$ to $(1, 1)$. I'm wondering how I can take the directional derivative in the direction of the vector from $(x,y)$ to $(1, 1)$. Let's say that the vector going from the origin through $(x,y)$ and through $(1, 1)$ is called $\beta$. How do I take the directional derivative in the direction of $\beta$ at the point $(x,y)$? Intuitively I would say that
$$
\beta\cdot\nabla v(x,y) = - u(x,y)
$$
but I'm not sure how I can show this rigorously.

Write $X$, $X_0$, and  $\underline 1$ for $(x,y)$, $(x_0,y_0)$ and $(1,1)$.   Unpacking the definition of the line integral $\int_{[X,\underline 1]}u ds$ of a scalar function (I assume $u$ is scalar but a similar calculation works for other cases) you want to differentiate $v(X)=|X-\underline 1|\int_0^{1} u(t\underline 1 + (1-t)X) dt $ in the direction $X_0=\underline1-X$. Note that the factor $|X-\underline 1|$ comes from the speed of the parameterisation I chose.  With no extra effort we can compute the derivative in all directions $X_0$. Setting $g(s)=|X +sX_0-\underline 1|$ and $f(s)=u(t\underline 1  + (1-t)(X + sX_0))$, you want the derivative of $g(s)\int^{1}_0 f(s,t)dt$ at $s=0$ which is $g'(0)\int f(0,t)dt +g(0)\int_0^1 \partial_s f(0,t)dt$.
$$g’(s)=\frac{X_0\cdot(X+sX_0-\underline1)}{|X+sX_0-\underline1|}$$ $$ f’(s)=X_0\cdot \nabla u(t\underline 1  + (1-t)(X + sX_0))$$
Put them together and you have the directional derivative at $X$ in direction $X_0$:
\begin{align}
 D_{X_0} v(X) &= \frac{X_0\cdot(X-\underline1)}{|X-\underline1|}\int_0^1 u(t\underline 1 +(1-t)X)dt \\&\quad + {|X-\underline1|}\int_0^{1}X_0\cdot\nabla u(t\underline 1  + (1-t)X ) dt
\\&= \frac{X_0\cdot(X-\underline1)}{|X-\underline1|^2}\int_{[X,\underline 1]} u ds + \int_{[X,\underline 1]} X_0\cdot \nabla u ds
\end{align}
Specialising now to $X_0=\underline 1-X$ one gets
$$ D_{\underline 1-X}v(X) = -  \int_{[X,\underline1]} u \,\text ds + \int_{[X,\underline1]} D_{\underline 1-X}u \,\text ds. $$
Since the answer is different from your guess, let me verify the answer  for a special case. If $u\equiv 1$ then $v(X)$ is just the unsigned distance of $X$ to $\underline 1$, $v(X) = |X-\underline 1|$. Its gradient at $X$ is $\frac{X-\underline 1}{|X-\underline 1|}$. Thus the directional derivative in the direction $\underline 1-X$ is $(\underline 1-X)\cdot \nabla v(X) = -|X-\underline 1|$ which is not $-u(X)=-1$.
However if you define the directional derivative in direction $X_0$ as $\frac{X_0}{|X_0|}\cdot\nabla $ then in this case one gets $-u(X)=-1$.
