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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?

user1999
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    Short answer: no. One is the second derivative, the other is the square of the derivative. – egreg Nov 26 '17 at 13:38
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    Try an example! Almost any $x(t)$ you look at other than the zero function will not have this two equal. – KCd Nov 26 '17 at 13:45
  • There is a family of functions for which the second derivative is the square of the first derivative: $f(x)=-\ln(ax+b)$, for arbitrary constants $a$ and $b$. But these are the only ones. I'm not sure but I don't think your manipulation is necessarily sound: The $d$ operator is a little weird. – Dan Uznanski Nov 26 '17 at 14:26

5 Answers5

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$$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt} \cdot \frac{dx}{dt}$$ whereas $$\frac{d^2x}{dt^2}=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{d\frac{dx}{dt}}{dt}$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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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 $t$

Dr. Sonnhard Graubner
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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 $t$

John_dydx
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No, they are not the same; but they can be the equal: https://youtu.be/wohyJ2heLso

Botond
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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.

amd
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