Understanding the derivation of the wave equation with gravity included

Question

I'm trying to understand this derivative of the wave equation with gravity included taken from here:

http://www.damtp.cam.ac.uk/user/examples/B8Lb.pdf

I understand that $\frac{dy}{dx}$ is the slope and drawing it out, we get

$$\frac{dy}{dx}=tan\theta$$

But I'm having troubles understanding the following (in the square):

How/where does this approximation come into play? I'm thinking it's something simple but it's just not coming to me

Answer 1

If you have a function you can calculate:

$$\frac{f(b)-f(a)}{b-a}\approx f'(a)$$ $$f(b)-f(a)\approx(b-a)f'(a)$$

Now take $f(x)=\frac{\partial y}{\partial x}$ and $b-a=\delta x$, you'll get $f''=\frac{\partial^2 y}{\partial x^2}$.