Find the directional derivative in the point $p$ in the direction $\vec{pp'}$

Question

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?

Answer 1

$$ D_{\vec{v_0}}f(\vec p)= \frac{\vec \nabla f(\vec p)\cdot (\vec {p}' - \vec p)}{||\vec {p}' - \vec p ||} $$

Answer 2

Hint

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

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