what does the second derivative tell us about?

It tells us the rate of change of the rate of change. For example, acceleration is the second derivative of a position function, like velocity is the first derivative.

why the second derivative operator is not d^2y/((d^2)(x^2)), i think this way is because the product of d/dx and dy/dx is d^2y/((d^2)(x^2))

It's just a (poor and confusing) convention, but when Leibnitz first invented this notation, he thought of units of physical quantities. For example, the second derivative ($\frac{d^2y}{dx^2}$) of position is acceleration. Acceleration has the units of $\frac{m}{s^2}$. And hence, the derivative (excluding the "d" part) is also $\frac{y}{x^2}$. There's another thing to consider that dx isn't d times x. It isn't a product and hence, dx $\cdot$ dx can't be $d^2x^2$ (d is an operator). But, we write $d^2$ in the numerator anyway, so this kinda invalidates it. Honestly speaking, this is the best explanation I could find. There's no reason why it couldn't have been $\frac{d^2y^2}{dx^2}$. People used $\frac{d^2y}{dx^2}$ and we got used to it.

is there such thing as third derivative?

Yes, we can find any number of derivatives as long as each derivative is also differentiable.

In problem #1. Why doesn't the 2 get multiplied against the entire derivative of cos(x/2)? It is only being multiplied by the first part of the chain: -sin(x/2). That doesn't seem correct.

It was multiplied by the derivative, that is why it went away. The derivative of 2cos(x/2) is 2 d/dx cos(x/2). You can use the chain rule to differentiate it, so you get 2*(1/2*-sin(x/2)). This simplifies to -sin(x/2) because 2 * 1/2 = 1. Does this help?

Main content

Course: AP®︎/College Calculus AB > Unit 3

Lesson 8: Calculating higher-order derivatives

Second derivatives review

Google Classroom

Review your knowledge of second derivatives.

What are second derivatives?

The second derivative of a function is simply the derivative of the function's derivative.

Let's consider, for example, the function

f (x) = x^{3} + 2 x^{2}

. Its first derivative is

f^{'} (x) = 3 x^{2} + 4 x

. To find its second derivative,

f^{″}

, we need to differentiate

f^{'}

. When we do this, we find that

f^{″} (x) = 6 x + 4

Want to learn more about second derivatives? Check out this video.

Notation for second derivatives

We already saw Lagrange's notation for second derivative,

f^{″}

Leibniz's notation for second derivative is

\frac{d^{2} y}{d x^{2}}

. For example, the Leibniz notation for the second derivative of

x^{3} + 2 x^{2}

\frac{d^{2}}{d x^{2}} (x^{3} + 2 x^{2})

Check your understanding

Problem 1

f (x) = 2 \cos (\frac{x}{2})

f^{″} (x) = ?

Problem 2

\frac{d^{2}}{d x^{2}} [\frac{10}{3 x^{3}}] = ?

Want to try more problems like this? Check out this exercise.

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zubair
Posted 5 years ago. Direct link to zubair's post “what does the second deri...”
what does the second derivative tell us about?
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(6 votes)
Answer
- JPOgle ✝
  Posted a year ago. Direct link to JPOgle ✝'s post “It tells us the rate of c...”
  It tells us the rate of change of the rate of change. For example, acceleration is the second derivative of a position function, like velocity is the first derivative.
  Button navigates to signup page
  (17 votes)
欧子
Posted a year ago. Direct link to 欧子's post “why the second derivative...”
why the second derivative operator is not d^2y/((d^2)(x^2)), i think this way is because the product of d/dx and dy/dx is d^2y/((d^2)(x^2))
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Venkata
  Posted a year ago. Direct link to Venkata's post “It's just a (poor and con...”
  It's just a (poor and confusing) convention, but when Leibnitz first invented this notation, he thought of units of physical quantities. For example, the second derivative ($\frac{d^2y}{dx^2}$) of position is acceleration. Acceleration has the units of $\frac{m}{s^2}$. And hence, the derivative (excluding the "d" part) is also $\frac{y}{x^2}$.
  
  There's another thing to consider that dx isn't d times x. It isn't a product and hence, dx $\cdot$ dx can't be $d^2x^2$ (d is an operator). But, we write $d^2$ in the numerator anyway, so this kinda invalidates it.
  
  Honestly speaking, this is the best explanation I could find. There's no reason why it couldn't have been $\frac{d^2y^2}{dx^2}$. People used $\frac{d^2y}{dx^2}$ and we got used to it.
  Comment on Venkata's post “It's just a (poor and con...”
  (9 votes)
Yu Aoi
Posted 3 years ago. Direct link to Yu Aoi's post “is there such thing as th...”
is there such thing as third derivative?
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(1 vote)
Answer
- KLaudano
  Posted 3 years ago. Direct link to KLaudano's post “Yes, we can find any numb...”
  Yes, we can find any number of derivatives as long as each derivative is also differentiable.
  Comment on KLaudano's post “Yes, we can find any numb...”
  (11 votes)
devore.wes
Posted 5 years ago. Direct link to devore.wes's post “In problem #1. Why doesn'...”
In problem #1. Why doesn't the 2 get multiplied against the entire derivative of cos(x/2)? It is only being multiplied by the first part of the chain: -sin(x/2). That doesn't seem correct.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- obiwan kenobi
  Posted 5 years ago. Direct link to obiwan kenobi's post “It was multiplied by the ...”
  It was multiplied by the derivative, that is why it went away. The derivative of 2cos(x/2) is 2 d/dx cos(x/2). You can use the chain rule to differentiate it, so you get 2*(1/2*-sin(x/2)). This simplifies to -sin(x/2) because 2 * 1/2 = 1. Does this help?
  Button navigates to signup page
  (6 votes)
Vanessa Slea
Posted 4 years ago. Direct link to Vanessa Slea's post “Given: dy/dx = x/y. Find...”
Given: dy/dx = x/y. Find the 2nd derivative d2y/dx2 in terms of x and y. I found (y^2 - x^2) / y^3 and it was marked as correct. Why then do certain calculators show the answer as
y/x ? Obviously, the calculator knows something I don't. What is it?
Button navigates to signup pageComment on Vanessa Slea's post “Given: dy/dx = x/y. Find...”
(2 votes)
Answer
tonymaione30
Posted 5 months ago. Direct link to tonymaione30's post “hey im struggling with un...”
hey im struggling with understanding something fundamental it seems. Im finding the derivative no problem but when i plug derivative in the numerator as a fraction using quotient rule i don't know how to evaluate it correctly what am i doing wrong?
eg
num- 6x^2*y-2x^3(2x^3/y) thanks for taking the time!
den- (y)^2
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(1 vote)
Answer
- cossine
  Posted 5 months ago. Direct link to cossine's post “So you have not mention w...”
  So you have not mention what you are trying to differentiate.
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  (1 vote)
Emily King
Posted 5 years ago. Direct link to Emily King's post “If I am asked to find f'(...”
If I am asked to find f'(x) of f(x)=x^4, does that mean that I am to find the second derivative (or f"(x))?
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(0 votes)
Answer
- Alex
  Posted 5 years ago. Direct link to Alex's post “If you are asked to find ...”
  If you are asked to find the first derivative (f'(x)), you are to find the first derivative (f'(x)).
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  (0 votes)