directional derivative, derivation of

Let $f(x,y)$ be a function where $x=x(t)$ and $y=y(t)$. Let $\overrightarrow{r}=a\widehat{𝐢}+b\widehat{𝐣}$ be the vector in the desired direction. The line through this vector is given parametrically by:

$x=x_0+at;y=y_0+bt$

The derivative of $f$ with respect to $t$ is as follows:

${\displaystyle \frac{\partial f}{\partial t}}={\displaystyle \frac{\partial f}{\partial x}}{\displaystyle \frac{dx}{dt}}+{\displaystyle \frac{\partial f}{\partial y}}{\displaystyle \frac{dy}{dt}}$

But from the equation of the line, we know that $\frac{dx}{dt}=a$ and $\frac{dy}{dt}=b$ so the derivative becomes:

${\displaystyle \frac{\partial f}{\partial t}}={\displaystyle \frac{\partial f}{\partial x}}a+{\displaystyle \frac{\partial f}{\partial y}}b=\nabla f\cdot \overrightarrow{r}$