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Second derivatives

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.F (LO)
,
FUN‑3.F.1 (EK)
,
FUN‑3.F.2 (EK)
Sal finds the second derivative of y=6/x². Second derivative is the derivative of the derivative of y.

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  • aqualine seedling style avatar for user Victoria
    We've learned so far that a derivative is the slope of a point at a graph; so does the second derivative represent something too?
    (43 votes)
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    • female robot grace style avatar for user G.
      The second derivative is the rate of change of the rate of change of a point at a graph (the "slope of the slope" if you will). This can be used to find the acceleration of an object (velocity is given by first derivative). You will later learn about concavity probably and the Second Derivative Test which makes use of the second derivative.
      (58 votes)
  • piceratops ultimate style avatar for user Joshua Ogunmefun
    I get that you aren't actually multiplying when you are finding the second derivative, but why is it the in the denominator of d^2/dx^2 the x gets the second power? why not the d or why not both?
    (12 votes)
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    • aqualine ultimate style avatar for user lnryoh
      I'm going to take a shot in the dark for sake of putting this out there: maybe it's because d approaches zero, and is the independent variable? Another thing I'll throw out there is if you nest two limits, the denominator ends up being squared. But I think it's most helpful to just realize that we generally don't pay attention why a radical looks like a radical, so we probably shouldn't put that much effort into critiquing the shape of the second derivative operator either.
      (5 votes)
  • aqualine ultimate style avatar for user Insatiable
    If I calculate the derivative of the second derivative, do I get the "third derivative"? Does this notion exist?
    (5 votes)
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  • piceratops seed style avatar for user Manuel Del Río Rodríguez
    Simple enough. Question is, what is the use of second derivatives? I know they probably have this-world applications that make them useful, but a priori, this just looks like an unecessary iteration of ... what? Rate of change of a rate of change of a rate of change...?
    (3 votes)
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    • starky ultimate style avatar for user Ron Joniak
      Yes, you said it! Rate of change. A classic example for second derivatives is found in basic physics. We know that if we have a position function and take the derivative of this function we get the rate of change, thus the velocity. Now, if we take the derivative of the velocity function we get the acceleration (the second derivative).

      Knowing the acceleration is crucially important for various physics applications. Thus, the second derivative is very useful. Many examples like this exist in various disciplines, it is highly important.
      (9 votes)
  • primosaur ultimate style avatar for user MathAddict3.14
    At , Sal says to apply the power rule. I applied the quotient rule, but we both got different answers. Why is that?
    (3 votes)
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    • duskpin sapling style avatar for user Vu
      y = 6/x²

      Quotient rule method. d/dx [u(x)/v(x)] = [u'(x)•v(x) - u(x)•v'(x)] / [v(x)]²
      But I remember it this way (it's equivalent, just rearranging the order) d/dx [u(x)/v(x)] = [v(x)•u'(x) - u(x)•v'(x)] / [v(x)]² so you will my calculation is in the order of the second formula.

      dy/dx (6/x²)
      = [(x²•0) - (6•2x)] / (x²)²
      = (0 -12x) / x⁴
      = -12x / x⁴
      = -12 / x³ (this is equivalent to -12x‾³ that Sal has)

      d²y/dx² (-12/x³)
      = [(x³•0) - (-12•3x²)] / (x³)²
      = [0 - (-36x²)] / x⁶
      = 36x² / x⁶
      = 36 / x⁴ (this is equivalent to 36x‾⁴)

      Is this what you have?
      (7 votes)
  • leaf grey style avatar for user Kshitij
    Just a thing I noticed.
    If we have a polynomial where all the powers are greater than 0(for example - x^3) if we keep taking the derivative then the function will eventually die(become 0).
    But, if instead, we have a polynomial where the powers are less than 0(for example - 1/x) if then we do the same pointless derivations the function will never die it will always be something.
    1/x
    -1/x^2
    2/x^3
    -6/x^4
    and so on.
    (3 votes)
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  • piceratops sapling style avatar for user McMahon
    Are there any videos that deal with taking higher derivatives implicitly? I just cannot figure out how to get to the second derivative using the first derivative (that I derived implicitly).
    (3 votes)
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    • leafers tree style avatar for user Donovan Snyder
      I don't see anything on here either, but to describe it quickly:
      Take the derivative of the first derivative, the same way you did the first time, where y'=(dy/dx).
      After you finish, replace any dy/dx's with what your answer was for the first derivative, and then simplify.
      (2 votes)
  • orange juice squid orange style avatar for user Rachel Oakley
    In one of the exercise problems for this vides, to get the first derivative of (10/3x^3), the answer says to take (10/3)*d/dx(1/x^3). Why is it not (10)*d/dx (1/3x^3)? This lead me to a different answer of (-10/9x^4) rather than (-10/x^4)
    (2 votes)
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    • female robot grace style avatar for user loumast17
      I think you may have made a mistake, I will go through both variations step by step.

      10 d/dx (1/3x^3)
      10 d/dx (3x^3)^-1
      10 (-(3x^3)^-2*9x^2)
      10(-9x^2/(9x^6))
      -10/x^4

      10/3 d/dx 1/x^3
      10/3 d/dx x^-3
      10/3 (-3x^-4)
      -10x^-4
      -10/x^4

      Let me know if something didn't make sense. With the first one it looks like you may have had a problem with the chain rule, but if that's not the case again feel free to respond.
      (2 votes)
  • leaf blue style avatar for user Zindagi :)
    According to that logic, shouldn't the second derivative be d^(2y/d^2x^2)?
    (2 votes)
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  • winston default style avatar for user Emma Gao
    Is the second derivative a nested function?
    (2 votes)
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Video transcript

- [Voiceover] Let's say that y is equal to six over x-squared. What I wanna do in this video is figure out what is the second derivative of y with respect to x. And if you're wondering where this notation comes from for a second derivative, imagine if you started with your y, and you first take a derivative, and we've seen this notation before. So that would be the first derivative. Then we wanna take the derivative of that. So we then wanna take the derivative of that to get us our second derivative. And so that's where that notation comes from. It likes you have a d-squared, d times d, although you're not really multiplying them. You're applying the derivative operator twice. It looks like you have a dx-squared. Once again, you're not multiplying 'em, you're just applying the operator twice. But that's where that notation actually comes from. Well, let's first take the first derivative of y with respect to x. And to do that, let's just remind ourselves that we just have to apply the power rule here, and we can just remind ourselves, based on the fact that y is equal to six x to the negative two. So let's take the derivative of both sides of this with respect to x, so with respect to x, gonna do that, and so on the left-hand side, I'm gonna have dy dx is equal to, now on the right-hand side, take our negative two, multiply it times the six, it's gonna get negative 12 x to the negative two minus one is x to the negative three. Actually, let me give myself a little bit more space here. So this negative 12 x to the negative three. And now, let's take the derivative of that with respect to x. So I'm gonna apply the derivative operator again, so the derivative with respect to x. Now the left-hand side gets the second derivative of y with respect to to x, is going to be equal to, well, we just use the power rule again, negative three times negative 12 is positive 36, times x to the, well, negative three minus one is negative four power, which we could also write as 36 over x to the fourth power. And we're done.