Question regarding the intuition for Directional Derivative Let's take a vector $\vec{v}$ = $a\hat{i}$+$b\hat{j}$ .
Let's take a function $f(x,y)$
The directional derivative along the  direction
$\hat{v}$=$\frac{a}{\sqrt{a²+b²}}\hat{i}$+$\frac{b}{\sqrt{a²+b²}}\hat{j}$ is,
$\nabla f$.$\hat{v}$=$\frac{a}{\sqrt{a²+b²}} \frac{\partial f}{\partial x}$+$\frac{b}{\sqrt{a²+b²}} \frac{\partial f}{\partial y}$.
Now my intuition is this ;

Let's say we need to find the rate of change of the function along some direction $\vec v$. (i.e.) $a\hat i+b\hat j=\vec v$
$\dfrac{\partial f}{\partial \vec v}\text{?}$
The rate of change in $x$ direction (i.e.) how much the function changes per unit length of $x$ is $\frac{\partial f}{\partial x}$.
Similarly the function changes by $\partial f$ per unit length in $y$ is $\frac{\partial f}{\partial y}$.
So per unit it has changed by $\frac{\partial f}{\partial x}$; if it has moved $a$ units, $a\frac{\partial f}{\partial x}$ is the change observed in the function by moving $a$ units. Similarly it has moved $b$ units in $y$ direction. Per unit length in $y$ direction its change is $\frac{\partial f}{\partial y}$ (at a pt.)
$b\frac{\partial f}{\partial y}$ is the change observed in the function by moving $b$ units in $y$ direction.
$\therefore \dfrac{\partial f}{\partial \vec v}=a\frac{\partial f}{\partial x}+b\frac{\partial f}{\partial y}$

Is my intuition right?
Why divide by its magnitude,shouldn't the derivative remain same ,regardless of the magnitude of vector ,because the direction remains constant ,if we observe a change in direction we can say that the slope is changing so is the derivative,but here the direction is the same ,so why the need to divide by magnitude?
 A: A vector and a direction are two different things. Remember what Despicable Me has taught you: vectors have both magnitude and direction. So to get just a direction, you have to factor out the magnitude.
For any non-zero vector $\vec v$, $$\hat v = \frac{\vec v}{\|\vec v\|}$$ is the unit vector pointing in the same direction as $\vec v$. Since unit vectors have their magnitude fixed, they can be used to represent directions only.
Thus if you want the derivative of $f$ with respect to a direction, you represent that direction with a unit vector, not an arbitrary vector.
Despite the name, a directional derivative is a derivative with respect to a vector, not just a direction. In particular, if $\mathbf a$ is a point, and $\vec v$ is a non-zero vector, $$\left.\dfrac{\partial f}{\partial \vec v}\right|_{\mathbf a} := \lim_{h\to 0}\dfrac{f(\mathbf a + h\vec v) - f(\mathbf a)}h$$ But note
$$\begin{align}\lim_{h\to 0}\dfrac{f(\mathbf a + h\vec v) - f(\mathbf a)}h &= \lim_{h\to 0}\dfrac{f(\mathbf a + h\|\vec v\|\hat v) - f(\mathbf a)}h\\&=\|\vec v\|\lim_{h\to 0}\dfrac{f(\mathbf a + h\|\vec v\|\hat v) - f(\mathbf a)}{\|\vec v\|h}\\&
=\|\vec v\|\lim_{u\to 0}\dfrac{f(\mathbf a + u\hat v) - f(\mathbf a)}u\end{align}$$
where $u = \|\vec v\|h$. Hence $$\left.\dfrac{\partial f}{\partial \vec v}\right|_{\mathbf a} = \|\vec v\|\left.\dfrac{\partial f}{\partial \hat v}\right|_{\mathbf a},\qquad\left.\dfrac{\partial f}{\partial \hat v}\right|_{\mathbf a} = \frac1{\|\vec v\|}\left.\dfrac{\partial f}{\partial \vec v}\right|_{\mathbf a}$$
So, to get the derivative with respect to a particular direction, you use a unit vector because the unit vector represents the direction only, nothing else. The derivative with respect to an arbitrary vector will be the derivative with respect to the direction multiplied by the magnitude of the vector.
