Q: What is another name for the third order derivative of displacement?

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We call "jerk" the third order derivative of position with respect to time, that is, the variation of acceleration. Some say that the derivative of jerk with respect to time (the fourth derivative of position with repsect to time) is called "jounce" or "snap".

in case of derivative w.r.t time first derivative with a variable x gives velocity second derivative gives acceleration thid derivative gives jerk

The first derivative of ln x is 1/x, which (for the following) you better write as x-1.Now use the power rule:Second derivative (the derivative of the first derivative) is -1x-2, the third derivative is the derivative of this, or 2x-3. You may now wish to write this in the alternative form, as 2 / x3.

1 divided by x to the third power equals x to the negative third. The derivative of x to the negative third is minus three x to the negative fourth.

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First derivative of displacement with respect to time = velocity. Second derivative of displacement with respect to time = acceleration. Third derivative of displacement with respect to time = jerk.

We call "jerk" the third order derivative of position with respect to time, that is, the variation of acceleration. Some say that the derivative of jerk with respect to time (the fourth derivative of position with repsect to time) is called "jounce" or "snap".

in case of derivative w.r.t time first derivative with a variable x gives velocity second derivative gives acceleration thid derivative gives jerk

The first derivative of ln x is 1/x, which (for the following) you better write as x-1.Now use the power rule:Second derivative (the derivative of the first derivative) is -1x-2, the third derivative is the derivative of this, or 2x-3. You may now wish to write this in the alternative form, as 2 / x3.

1 divided by x to the third power equals x to the negative third. The derivative of x to the negative third is minus three x to the negative fourth.

The 3rd derivative is very useful in the process of trying to findthe maximum and minimum points of the 2ndderivative.

Can you be more specific.

The first derivative is m and the second is 0 so the third is also 0.

If the second derivative of a function is zero, then the function has a constant slope, and that function is linear. Therefore, any point that belongs to that function lies on a line.

You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.

The derivative of y = 1/3 x3 - 3x2 + 8x + 1/3 is x2 - 6x + 8. You can determine this for yourself by the rules. The derivative of a constant (e.g. 1/3) is 0. The derivative of xn for positive n (actually all nonzero n) is nxn-1. And if the derivative of f(x) is f'(x), then the derivative of k f(x) is k f'(x). Put all these together and you get the above result.

"Avait" is the derivative of the verb "Avoir" in french, meaning "to have". It is third-person imperfect. "Il avait" - "He had"