Find the directional derivative in the point $p$ in the direction $\vec{pp'}$ Find the directional derivative in the point $p$ in the direction $\vec{pp'}$. In other words, if $\vec{v_0}$ is the unitary vector in the direction of $\vec{pp}$. you need find $D_{\vec{v_0}}f(p)$.

$f(x,y,z)=x^2+3xy+y^2+z^2$ with $p=(1,0,2)$ and $p'(-1,3,4)$

I dont understand the exercise, is confused his statement. Can someone explain me how solve this type of exercise?
Hint

If you don’t know what the gradient is, you should use directly the limit of the directional derivative. Which means


Compute $v_0 = \frac{pp^\prime}{\Vert pp^\prime\Vert}= (a,b,c)$
Compute the limit
$$\lim\limits_{h \to 0} \frac{f(1+a.h,0+b.h,2+c.h)-f(1,0,2)}{h}$$
$$ D_{\vec{v_0}}f(\vec p)= \frac{\vec \nabla f(\vec p)\cdot (\vec {p}' - \vec p)}{||\vec {p}' - \vec p ||}   $$