If v = dx/dt, then how to calculate a = d²x/dt²?

[Indranil]

If v = dx/dt, then how to calculate a = d²x/dt²?

As we know v = dx/dt and a = dv/dt,
so a = d²x/dt² but how to get a = d²x/dt² by putting the velues in the a = dv/dt. I have done below;
a = dv/dt = d(dx/dt) /dt = d.dx/dt times 1/dt = d²x/d²t² but it should be d²x/dt². Could you please find out my mistakes here?

 

[MarkFL]

$\displaystyle \displaystyle a=\frac{dv}{dt}= \frac{d}{dt}\left(\frac{dx}{dt}\right)= \frac{d^2x}{dt^2}$

 

[mmm4444bot]

… a = dv/dt = d(dx/dt) /dt = d.dx/dt times 1/dt = d²x/d²t² …

It seems like you misunderstand some calculus notation. Symbol d is not a number. We don't multiply d.dx

When you see the symbol d/dt in front of a function, it represents a derivative operator, not a number. In other words, the notation d/dt means "take the derivative with respect to t" of the function written after it.

In the expression d/dt (dx/dt)

the symbol d/dt means "take the derivative" and the symbol dx/dt means "the first derivative of the displacement function x(t)"

Therefore, the expression d/dt (dx/dt) tells us to take the derivative of a derivative. The result is the second derivative of the function x(t).

d^2x/dt^2 is a notation for the second derivative of x(t). :cool:

 

[Indranil]

It seems like you misunderstand some calculus notation. Symbol d is not a number. We don't multiply d.dx

When you see the symbol d/dt in front of a function, it represents a derivative operator, not a number. In other words, the notation d/dt means "take the derivative with respect to t" of the function written after it.

In the expression d/dt (dx/dt)

the symbol d/dt means "take the derivative" and the symbol dx/dt means "the first derivative of the displacement function x(t)"

Therefore, the expression d/dt (dx/dt) tells us to take the derivative of a derivative. The result is the second derivative of the function x(t).
d^2x/dt^2 is a notation for the second derivative of x(t). :cool:

Then what will first derivative of x(t)

 

[Indranil]

$\displaystyle \displaystyle a=\frac{dv}{dt}= \frac{d}{dt}\left(\frac{dx}{dt}\right)= \frac{d^2x}{dt^2}$

If d(dx) = d²x then why not possible dt(dt) = d²t²? Still I don't understand. Please simplify.

 

[Dr.Peterson]

If d(dx) = d²x then why not possible dt(dt) = d²t²? Still I don't understand. Please simplify.

This is entirely a matter of notation -- how we choose to write the second derivative. Ultimately, you just have to learn that this is how it is written.

But the motivation is that, using $\displaystyle \displaystyle \frac {d}{dx}$ to represent the operation of taking the derivative of a function, the second derivative is represented as $\displaystyle \displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)$, and we write that as if d were a number, even though it isn't: $\displaystyle \displaystyle \frac{ddy}{dx dx} = \frac{d^2y}{(dx)^2}= \frac{d^2y}{dx^2}$. We don't bother to use parentheses around the dx because in some sense the d, as a "differential operator", is thought of as tightly bound to what follows it; and we don't distribute and write d²x² because that would suggest we are doing something to x², which we are not.

Another way to look at it is in terms of the definition. The derivative is a limit of a fraction, $\displaystyle \displaystyle\frac {dy}{dx} = lim_{\Delta x\to 0}\frac {\Delta y}{\Delta x}$, so the second derivative is the limit of $\displaystyle \displaystyle \frac {\Delta \frac {\Delta y}{\Delta x}}{\Delta x}$, and this can almost be thought of as $\displaystyle \displaystyle \frac {\Delta(\Delta y)}{(\Delta x)^2}$. Here, again, $\displaystyle \Delta$ is not a number, but an operator, so distribution is not valid.

But, again, this is simply a notation that reminds us of these relationships, and happens to work well; if you try to analyze it too closely, it falls apart. Just accept it as the way we write these things.

 

[mmm4444bot]

Then what will first derivative of x(t)

dx/dt represents the first derivative of x(t).

d/dt [x(t)] yields dx/dt

 

[HallsofIvy]

Then what will first derivative of x(t)

That depends upon what x is! If, for example, $\displaystyle x= t^3+ 3t+ 4$ then $\displaystyle v=\frac{d(t^3+ 3t+ 4)}{dt}= 3t^2+ 3$ and then $\displaystyle a= \frac{dv}{dt}= \frac{d(3t^2+ 3)}{dt}= 6t$.

Or if $\displaystyle x= sin(\omega t)$ then $\displaystyle v= \frac{d(sin(\omega t)}{dt}= \omega cos(\omega t)$ and then $\displaystyle a= \frac{dv}{dt}= -\omega^2 sin(\omega t)$.