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

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The expression $\frac{d}{dx} \frac{dy}{dx}$ represents the second derivative of y with respect to x, commonly denoted as $\frac{d^2y}{dx^2}$. For the function $y = t^2$, where $x = t$, the first derivative is $\frac{dy}{dx} = 2t$, and the second derivative is $\frac{d^2y}{dx^2} = 2$. When considering the function $y = t^2 + 2t + 1$, the same process applies, yielding a second derivative of 2. The discussion also highlights the relationship between derivatives and physical concepts such as acceleration and velocity, emphasizing the importance of rewriting functions in terms of x for clarity.

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Can someone please tell me what is meant by ##\frac{d}{dx} \frac{dy}{dx})##?
 
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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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