Change in output of $f(x,y)$ when $hv$ added to $x$

Question

From Khan Academy: change in output of $f$

How does adding the $hv_1$ to $x$ result in change of output of $f$ to:

$hv_1(\frac{df}{dx(x,y)})$ ? I made a sample function $f(x,y)$ = $xy$ and found partial derivatives with respect to $x$ is $y$ and with respect to $y$ is $x$ but connecting the idea of a partial derivative to the output of a function is where I am lost.

Answer 1

I wanted to delete this but I received help with editing the question so I am posting an answer.

Working out f(x0 + hv1, yo) when f(x0, y0) = x0 * y0

I get x0y0 + hv1y0.

Since df/dx = y0, the output of the function f increased by hv1 * y0 which is hv1(df/dx).

But the Khan writer makes it sound like there is a faster more analytical or formulatic way of stating this rather than working out the equation and then connecting the dots.

So I am still confused.