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### Course: Differential Calculus>Unit 3

Lesson 9: 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 $f\left(x\right)={x}^{3}+2{x}^{2}$. Its first derivative is ${f}^{\prime }\left(x\right)=3{x}^{2}+4x$. To find its second derivative, ${f}^{″}$, we need to differentiate ${f}^{\prime }$. When we do this, we find that ${f}^{″}\left(x\right)=6x+4$.
Want to learn more about second derivatives? Check out this video.

## Notation for second derivatives

We already saw Lagrange's notation for second derivative, ${f}^{″}$.
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 $\frac{{d}^{2}}{d{x}^{2}}\left({x}^{3}+2{x}^{2}\right)$.

## Check your understanding

Problem 1
$f\left(x\right)=2\mathrm{cos}\left(\frac{x}{2}\right)$
${f}^{″}\left(x\right)=?$

Problem 2
$\frac{{d}^{2}}{d{x}^{2}}\left[\frac{10}{3{x}^{3}}\right]=?$

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

## Want to join the conversation?

• 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?
(1 vote)
• 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.
(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?
• 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?
• 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)
• 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))?