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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.

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You are correct.

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

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You are correct, the book is wrong. No you can't differentiate by different variables. You 1, book 0.

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