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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 3

Lesson 8: Calculating higher-order derivatives# 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.

## Want to join the conversation?

- 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?(44 votes)
- 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.(61 votes)

- 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?(13 votes)
- 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.(6 votes)

- If I calculate the derivative of the second derivative, do I get the "third derivative"? Does this notion exist?(6 votes)
- Yes, you get the third derivative. You can differentiate a (typical, well-behaved) function as many times as you want to get a fourth derivative, fifth derivative, or any nth derivative.(8 votes)

- 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...?(4 votes)
- 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.(10 votes)

- At1:04, Sal says to apply the power rule. I applied the quotient rule, but we both got different answers. Why is that?(4 votes)
- 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?(8 votes)

- 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.(4 votes)- All rational expressions with negative and/or non-integer powers will have infinitely many non-trivial derivatives. (I don't mean to argue over semantics, but an expression with negative powers is not a polynomial, it is rational.)(5 votes)

- 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).(4 votes)
- 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.(3 votes)

- 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)(3 votes)
- 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.(3 votes)

- According to that logic, shouldn't the second derivative be d^(2y/d^2x^2)?(2 votes)
- Is the second derivative a nested function?(2 votes)

## 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.