# Sign inconsistency in finding higher order partial derivates of $y=f(x-vt)$

For the function $$y=f(x-vt)$$ where $$x$$ and $$t$$ are variables and $$v$$ is a constant, I was given the following options where I'm allowed to choose more than one:

(a) $$\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}$$

(b) $$\frac{\partial y}{\partial t}=-v\frac{\partial^2 y}{\partial x^2}$$

(c) $$\frac{\partial^2 y}{\partial t^2}=-v^2\frac{\partial^2 y}{\partial x^2}$$

(d) $$\frac{\partial^2 y}{\partial t^2}=v^2\frac{\partial^2 y}{\partial x^2}$$

This is how I approached:

We're given that $$y=f(x-vt)$$. So,

\begin{align} \frac{\partial y}{\partial t}&=-v f'(x-vt)\tag 1\\ \frac{\partial y}{\partial x}&=f'(x-vt)\tag 2 \end{align}

Using $$(2)$$ in $$(1)$$, we get:

$$\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}\tag 3$$

which is same as option (a) and according to my book it's one of the correct options.

Next for $$\frac{\partial^2 y}{\partial t^2}$$, I took the corresponding partial derivatives of $$(1)$$ and $$(2)$$ to arrive at $$(4)$$ and $$(5)$$ as follows:

\begin{align} \frac{\partial^2 y}{\partial t^2}&=v^2 f''(x-vt)\tag 4\\ \frac{\partial^2 y}{\partial x^2}&=f''(x-vt)\tag 5 \end{align}

Using $$(5)$$ in $$(4)$$, I got:

$$\frac{\partial^2 y}{\partial t^2}=v^2\frac{\partial^2 y}{\partial x^2}\tag 6$$

which is same as option (d). However, my book says the answer is (c) $$\frac{\partial^2 y}{\partial t^2}=\color{red}{-}v^2\frac{\partial^2 y}{\partial x^2}$$ (with a negative sign). I haven't spotted any errors in my method. It would be helpful if you could explain where I went wrong and how to arrive at the correct answer.

In fact, $$f(x-vt)$$ is a solution of what is known as the wave equation, which is the partial differential equation $$\frac{\partial ^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$$.