differentiating formula A particle’s motion is described by the following equation: $x = A \cos \cos (ωt)$ . The probability that the particle can be found
between position $x$ and $x + dx$ is $dp$ $p$ What is the expression for $\frac{dp}{dx}$?

I have a value given.
$$x=A \cos \cos (ωt)$$
The equation was looking weird for me. That's why I thought that, that's why I thought that is $cos^2$. And, the question is saying that $dp=x+dx$, What is the expression for $\frac{dp}{dx}$?
$$x=A\cos\cos(\omega t)$$
$$dx=A\omega \sin \sin(\omega t)$$
$$\frac{dp}{dx}=\frac{x+dx}{dx}=\frac{A\cos\cos(\omega t)+A\omega \sin sin(\omega t)}{A\omega \sin \sin (\omega t)}=\frac{A(\cos^2(\omega t)+\omega \sin^2(\omega t)}{A \omega \sin^2(\omega t)}$$
What the next expression should look like? I think I am doing something wrong with differentiation.
I am assuming you are trying to solve the following problem:
The position of a particle is given by the formula : $x = A cos^2(wt)$ and we want to find the derivative with respect to the position of $p(x)$ which is the probability density of finding a particle at a certain position.
As explained here, the probability that the oscillating particle is at a given value of x is simply the fraction of time that it spends there, which is inversely proportional to its velocity in this point. The particle must stop completely (for a moment) before reversing its direction, and so it spends the most time where the spring is either fully compressed or fully extended. It spends the least time where its velocity is greatest i.e., where the spring is at its equilibrium length.
$v(t)=\frac{dx(t)}{dt} = -2Acos(wt)sin(wt)w = -Asin(2wt)w$
$p(x)$ is proportional to $\frac{1}{v(t)}$ at $t=\frac{1}{w} \arccos{\sqrt \frac{x}{A}}$
Then you can normalise $p(x)$ and take the derivative of this expression to find $\frac{dp}{dx}$. But in my opinion taking the derivative of $p$ with respect to time doesn't make any sense.
I am assuming you are trying to solve the following problem:
The position of a particle is given by the formula : $x = A cos^2(wt)$ and we want to find the derivative with respect to the position of $p(x)$ which is the probability density of finding a particle at a certain position.
As explained here, the probability that the oscillating particle is at a given value of x is simply the fraction of time that it spends there, which is inversely proportional to its velocity in this point. The particle must stop completely (for a moment) before reversing its direction, and so it spends the most time where the spring is either fully compressed or fully extended. It spends the least time where its velocity is greatest i.e., where the spring is at its equilibrium length.
$v(t)=\frac{dx(t)}{dt} = -2Acos(wt)sin(wt)w = -Asin(2wt)w$
$p(x)$ is proportional to $\frac{1}{v(t)}$ at $t=\frac{1}{w} \arccos{\sqrt \frac{x}{A}}$
Then you can normalise $p(x)$ and take the derivative of this expression to find $\frac{dp}{dx}$. But in my opinion taking the derivative of $p$ with respect to time doesn't make any sense.