What Is the Meaning of $\frac{d}{dx} \frac{dy}{dx}$?

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Discussion Overview

The discussion centers on the interpretation and calculation of the expression ##\frac{d}{dx} \frac{dy}{dx}##, specifically in the context of derivatives in calculus. Participants explore its meaning, applications, and implications in various scenarios, including parametric equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Some participants assert that ##\frac{d}{dx} \frac{dy}{dx}## represents the second derivative of y with respect to x.
  • One participant provides an example with ##y = t^2## and ##x = t##, calculating ##\frac{dy}{dx} = 2t##.
  • Another participant suggests rewriting ##y = t^2## as ##y = x^2## to find the second derivative, indicating a method for simplification.
  • Participants discuss a different function, ##y = t^2 + 2t + 1##, and inquire about its second derivative, indicating a continued exploration of the concept.
  • One participant relates the second derivative to physical concepts such as acceleration, velocity, and higher derivatives, introducing a contextual application of the mathematical concept.
  • Several participants suggest using parametric forms to express derivatives, indicating a preference for this approach in certain cases.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of ##\frac{d}{dx} \frac{dy}{dx}## as the second derivative, but there are varying approaches to calculating it, particularly regarding the use of parametric equations. The discussion remains exploratory with no consensus on a single method or interpretation.

Contextual Notes

Some participants mention rewriting functions in terms of x or t, indicating that the choice of variable may affect the approach to finding derivatives. There is also a lack of resolution on the best method for parametric derivatives.

askor
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Can someone please tell me what is meant by ##\frac{d}{dx} \frac{dy}{dx})##?
 
Last edited:
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That's the 2nd derivative of y with respect to x.
 
If ##y = t^2## and ##x = t##, what is its ##\frac{d}{dx} (\frac{dy}{dx})##?

I know that ##\frac{dy}{dx} = \frac{2t}{1} = 2t##.
 
You can rewrite ##y = t^2## as ##y = x^2##. Then ##\frac{dy}{dx} = 2x##.

Take the next derivative of that.
 
What about, if, ##y = t^2 +2t + 1## and ##x = t##, what is its 2nd derivative?
 
Same idea. Rewrite it in terms of x and take the derivatives.
 
It looks like acceleration to me ##-## 2nd derivative of position with respect to time, the 1st being velocity, the 3rd being jerk, the 4th, jounce or snap (jounce being by some called snap at least partly so that some whimsical people can wryly call the 4th, 5th, and 6th, snap crackle, and pop, respectively.
 
Doc Al said:
Same idea. Rewrite it in terms of x and take the derivatives.
Can you show in terms of t (parametric)? I know that ##\frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt}##.
 
Just use t as the variable and take derivatives with respect to t.
 
  • #10
Can you show in form of t (parametric)? I know that ##\frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt}##.
 

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