What does $\frac{d^2 u}{dt^2}$ mean?

Question

When it comes to taking a derivative, what does $\displaystyle \frac{d^2 u}{dt^2}$ mean ? Does it mean taking derivative of the function twice with respect to $t$. If yes, why is then $d^2 u$ squared? Thanks in advance!

Answer 1

It means to take $\frac{du}{dt}$ twice.

Answer 2

That means the second derivative of $u$ with respect to $t$. The "squaring" you see is a notation convention. Since the second derivative of $u$ can be obtained from applying the differential operator twice, putting the "square" on the $d$ is shorthand for $$\frac{d}{dt}\frac{d}{dt}u$$ or equivalently, $$\frac{d^2 u}{dt^2}$$

Answer 3

this simply means it is the second derivative of u with respect to t

Answer 4

The $d$ represents the differential operator which is applied twice because we are wanting to find the second derivative. Now, $dt$ is thought of as a change in $t$, which is a quantity. We can think of it like this $$\frac{d}{dt}\left(\frac{du}{dt}\right)=\frac{d(du)}{(dt)^2}=\frac{d^2u}{dt^2}$$

Answer 5

This basically means that taking the second derivative of any given function. To elaborate this you can consider some function $$y=f(t)$$ Now lets find its derivative $$y'= \frac{d(f(t))}{dt}$$. Now lets say I want to know what would the derivative of the earlier function $y'$ would be. That will be $$y''=\frac{d(\frac{d(f(t))}{dt})}{dt}$$ and this would be $$y''=\frac{d^2u}{dt^2}$$ where $u=f(t)$

Now in terms of physics you can think this as if i have the position as the function of time if i find its derivative it gives me the velocity and again if i find its derivative it gives me the acceleration . So the second derivative of the position gives me the acceleration

Answer 6

It might seem odd that there are two occurrences of $2$, right? Why do we see a $2$ in both the numerator and denominator?

$$\frac{d^2 u}{dt^2}$$

Consider the top number to represent the number of derivatives. Here we see there are two derivatives being performed on the function $u$.

$$\frac{d}{dt} \frac{d}{dt}u$$

The bottom number though, represents what variables we're deriving in respect to. In your case, it's $t$ in both occurrences. It's possible, though, to take a derivative with respect to, say, $x$ and then with respect to another variable, say $y$.

$$\frac{\partial^2 u}{\partial{y}\partial{x}}$$

Above, we still are taking two (partial) derivatives, and the denominator tells us this is for two variables in the order $x$ then $y$.