Are $\left(\frac{dx}{dt}\right)^2$ and $\frac{d^2 x}{dt^2}$ same things? Is the second derivative of x(t) with respect to t equal to the square of the first derivative of x(t) with respect to t? In other words is the following correct:
$$\frac{d^2x}{dt^2} = \left(\frac{dx}{dt}\right)^2 = \frac{dx}{dt}\times\frac{dx}{dt}$$
edit:
The reason I ask is the following issue I encountered:
$\frac{dx}{dt}\times{di} + {dx} = 0$    Equation 1.
if I multiply the above equation by $(\frac{1}{dt})$
would the following be correct?:
$(\frac{dx}{dt})\times(\frac{di}{dt})+ \frac{dx}{dt} = 0$
What Im confused is above I just multiplied the Equation 1 with $(\frac{1}{dt})$. Is multiplying the Equation 1 this way different than taking the derivative of the Equation 1 wrt t?
 A: $$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt} \cdot \frac{dx}{dt}$$ whereas $$\frac{d^2x}{dt^2}=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{d\frac{dx}{dt}}{dt}$$
A: the term $$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt}\cdot \frac{dx}{dt}$$
and the term $$\frac{d^2x}{dt^2}$$ means the second derivative of $x(t)$ with respect to $t$
A: They are not the same 
$$\left(\frac{dx}{dt}\right)^2 = \frac{dx}{dt} \times \frac{dx}{dt}$$ 
$$\left(\frac{dx}{dt}\right)^2$$ is the square of the derivate of $x(t)$ with respect to $t$
$$ \frac{d^2x}{dt^2}$$ is the second derivate of $x(t)$ with respect to $t$
A: No, they are not the same; but they can be the equal: https://youtu.be/wohyJ2heLso
A: A dimensional analysis should suggest that they are not the same thing at all: Suppose we measure $x$ in meters and $t$ in seconds. The first derivative ${dx\over dt}$ then has units of meters per second, and its square $\left({dx\over dt}\right)^2$ therefore has units of square meters per second squared. On the other hand, the second derivative ${d^2x\over dt^2}={d\over dt}\left({dx\over dt}\right)$ has units of meters per second squared. So, even though they might be numerically equal in some cases, they measure completely different things.
