You can't use an area relation because we're not told what $\dfrac{dA}{dt}$ is. If you draw out the situation, we see that we can actually use a different relation to solve this problem.
The equation that you want to use comes directly from the Pythagorean theorem; in particular, if $x$ is the distance the base of the ladder is from the wall and $y$ is the the height of the ladder along the wall, then for the 5 m long ladder, we have that $x^2+y^2 = 25$.
Differentiating this will then give you the appropriate equations involving $\dfrac{dx}{dt}$ (i.e. the rate the ladder is sliding away from the wall) and $\dfrac{dy}{dt}$ (i.e. the rate the ladder is sliding down the wall). You can then plug in all the known values at the appropriate instant in time (when $x=3$, $y=\ldots$ [which can be found using the Pythagorean theorem], and $dx/dt = 0.4\text{ m/s}$) and then solve for $\dfrac{dy}{dt}$.
Hopefully this is enough information to help you get the correct solution. :-)
Well, $\theta$ is in degrees, so the derivative of $\sin(\theta)$ is not $\cos(\theta)$ but $\frac{\pi}{180}\cos(\theta)$. This is because:
$$\frac{d}{dx}[\sin(\theta)] = \cos(\theta)$$ for radian values of $\theta$ only.
To get degree values:
$\frac{d}{dx}[\sin(\theta)]$ with $\theta$ in radians is equal to $\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)]$ with theta in degree values but the sine function still taking radian values. Therefore, the new derivative is:
$$\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)] = \frac{\pi}{180}\cdot \cos(\frac{\pi}{180}\cdot \theta)$$
where theta is in degrees but cosine is still taking radian values. Therefore, if we change cosine and sine to take degree values we can simplify to:
$$\frac{d}{dx}[\sin(\theta)] = \frac{\pi}{180}\cdot \cos(\theta)$$ for degree values of $\theta$
Therefore, with your previous evaluation, I believe the correct answer is actually $\frac{\pi}{36}$. If I assume that as your answer, I can answer your other questions:
For your first question, we can be sure of that because $x'$ is equal to: $\frac{dx}{d\theta}$, which is read as the rate in change of $x$ with respect to $\theta$.
For your second question, $x$ is already in feet and $\theta = 60$ degrees (as stated in the question), so all you have to do is state it like this:
The rate of change of $x$ with respect to $\theta$ at $\theta$ = 60 degrees is $\frac{\pi}{36}$ feet per degree.
Best Answer
Whenever I've encountered the phrase "rate of change of the distance", it has just meant the derivative of the distance with respect to the independent variable. As I recall, the independent variable has always been time, like it is in your particular question, so the phrase is referring to the speed, i.e., $s = \frac{d(\text{distance})}{dt}$.
Here the distance means the the length of the straight line, that is perpendicular to the wall, from the wall to the bottom of the ladder. As my answer says, the rate of change of distance is just the derivative of this distance function wrt time, i.e., the speed. In simpler phrasing, it just means how fast is the bottom of the ladder moving away from the wall.