Main content

### Course: AP®︎ Calculus BC (2017 edition) > Unit 4

Lesson 2: Second derivatives# Second derivatives review

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 ${f}^{\prime}(x)=3{x}^{2}+4x$ . To find its second derivative, ${f}^{\u2033}$ , we need to differentiate ${f}^{\prime}$ . When we do this, we find that ${f}^{\u2033}(x)=6x+4$ .

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

## Notation for second derivatives

We already saw Lagrange's notation for second derivative, ${f}^{\u2033}$ .

Leibniz's notation for second derivative is $\frac{{d}^{2}y}{d{x}^{2}}$ . For example, the Leibniz notation for the second derivative of ${x}^{3}+2{x}^{2}$ is $\frac{{d}^{2}}{d{x}^{2}}}({x}^{3}+2{x}^{2})$ .

## Check your understanding

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

## Want to join the conversation?

- what does the second derivative tell us about?(7 votes)
- 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.(20 votes)

- 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))(4 votes)
- 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.(9 votes)

- is there such thing as third derivative?(1 vote)
- Yes, we can find any number of derivatives as long as each derivative is also differentiable.(11 votes)

- 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.(1 vote)
- 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?(6 votes)

- 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?(2 votes) - 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(1 vote)- So you have not mention what you are trying to differentiate.(1 vote)

- I have that \dfrac{dy}{dx} = e^{5y} and I want to find \dfrac{d^2y}{dx^2}.

So, I solve to here:

\dfrac{d^2y}{dx^2}=5e^{5y}\cdot\dfrac{dy}{dx}

I can substitute in e^5y for \dfrac{dy}{dx}.

What happens if instead I divide both sides by \dfrac{dy}{dx}?(1 vote) - 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))?(0 votes)
- If you are asked to find the first derivative (f'(x)), you are to find the first derivative (f'(x)).(0 votes)