Deriving the equation of motion for transverse vibrations on a taut string. Please help explain how one expression leads to the next.

Question

The image shows a section of my textbook, I am teaching myself how to derive the wave equation describing transverse vibrations on a taut string.

I can't work out how the expression in the square brackets equates to the second partial derivative w.r.t. x multiplied by dx.

I am familiar (but clearly not familiar enough!) with the difference quotient (or as the textbook calls it, the differential coefficient), but I am not able to grasp how the expression inside the brackets equates to the second partial derivative. I have tried expressing each term as a separate difference quotient but I can't get to the second expression that appears on the image.

I would really appreciate some help on this, thanks.

Answer 1

That's just the definition: let's say $$f(x)=\left(\frac{dy}{dx}\right)_x$$Then $$f(x+dx)=\left(\frac{dy}{dx}\right)_{x+dx}$$and the derivative of $f$ with respect to $x$ is $$\frac{df}{dx}=\frac{f(x+dx)-f(x)}{dx}$$ Plugging back what $f$ is, you get $$\frac{d^2y}{dx^2}=\frac{\left(\frac{dy}{dx}\right)_{x+dx}-\left(\frac{dy}{dx}\right)_{x}}{dx}$$