# What is the Higher Derivative of a function?

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1. Nov 15, 2015

I'm relatively new to calculus and I have a new chapter in my study which is on the Implicit Function, Implicit Differentiation and Higher Derivatives of a function, the problem is I don't understand the meaning of a 2nd or 3rd or whatever the higher derivative of a function is, what I know is that a derivative of a function is its average rate of change , so I want a simple explanation yet a comprehensive one on what's a higher derivative of a function .

2. Nov 15, 2015

### Orodruin

Staff Emeritus
You need to revise your understanding of what te first derivative is as well. It is not the average rate of change, it is the rate of change at a given point.

3. Nov 15, 2015

Thanks for your reply , I think I'm right, as I didn't define the derivative of a function at a specific point, let f(X)=X2 then the average rate of change of F(X) = 2X correct me if I'm wrong .

4. Nov 15, 2015

### Orodruin

Staff Emeritus
You are wrong. 2x is the rate of change at x, not the average rate of change.

5. Nov 15, 2015

Then tell me what is the average rate of change in general and in case of the function that I previously mentioned besides answering my main question .

6. Nov 15, 2015

### Orodruin

Staff Emeritus
The average rate of change depends on which points you consider the average between. The rate of change at a point, ie, the derivative, is defined as the limit of this quantity when the interval goes to zero.

7. Nov 15, 2015

### Erland

Suppose you drive a distance of 100 miles in 2 hours, at irregular speed. Sometimes you drive fast, sometimes slow. The average rate of change of the distance covered during these two hours is the same as the average speed, 100/2 = 50 miles per hour.
But you don't drive 50 mi/h all the time, your speed is irregular. What the speedometer in your car shows is the instantaneous speed at each moment. This instantaneos speed is the same as the (instantaneous) rate of change of the distance covered, and this is the same as the derivative of the distance with respect to the time. This derivative / rate of change / instantaneous speed varies. It could for example be 60 mi/h at t = 1 h, but 40 mi/h at t = 1,5 h, etc - it is what the speedometer shows at each moment in time.

8. Nov 15, 2015

Thanks, I appreciate your effort, now I understand what a derivative of a "Relation" is , but still , no one has answered my question yet , what is a higher derivative of a function , can you please illustrate it on that example you mentioned , thanks again .

9. Nov 15, 2015

### Krylov

Are you talking about functions defined on open subsets of $\mathbb{R}^n$ for $n = 1$ or for $n > 1$? Your question is easier to answer in the former case than in the latter case.

10. Nov 15, 2015

### zinq

The original poster is just beginning to understand what a derivative is. (In one dimension.)

It would be better for him to definitely understand the concept and how to use it before going on to try to understand higher derivatives.

11. Nov 15, 2015

### Erland

The second derivative is the derivative of the derivative, or the rate of change of the rate of change. So, since speed is the (first) derivative of distance, the second derivative of distance is the (first) derivative of speed, that is, acceleration.Then the third derivative is the derivative of the second derivative, etc.

12. Nov 15, 2015

You are right , but I have nothing to do with it , as I live in Egypt and the curriculum is so dumb in every aspect ,but I must be good at it to pass the exams with good grades and go to the faculty of engineering .

13. Nov 15, 2015

Thank you so much , your replies really helped me , especially when you gave an example , but what does it mean if the average rate of change function in F(x) at x is M(h) = Δy/Δx where Δy = f(x+h)-f(x) , Δx = h , has a limit at h→0 then this limit is called the instantaneous rate of change (first derivative) of y at x , and why the limit is when h→0 and not anything else , thanks in advance .

14. Nov 15, 2015

Actually , I don't know what does $\mathbb{R}^n$ for for $n > 1$ mean , but I know that $\mathbb{R}^n$ for $n = 1$ means that the function is defined on subsets of "real numbers" correct me if I'm wrong , thanks .

15. Nov 15, 2015

### HallsofIvy

The first derivative of the function f is the instantaneous rate of change of f. The second derivative is the instantaneous rate of change of the first derivative, the third derivative is the instantaneous rate of change of the second derivative, etc. That's all there is to it.

Last edited by a moderator: Nov 15, 2015
16. Nov 16, 2015

Thanks for your reply , but can you please answer my question "what does it mean if the average rate of change function in F(x) at x is M(h) = Δy/Δx where Δy = f(x+h)-f(x) , Δx = h , has a limit at h→0 then this limit is called the instantaneous rate of change (first derivative) of y at x , and why the limit is when h→0 and not anything else" thanks in advance .

17. Nov 16, 2015

I'm talking in general .

18. Nov 16, 2015

### Krylov

You are right about $n = 1$. For $n > 1$ it means that $f$ is a function of $n$ variables, but depending on the "open subset" on which it is defined, not all variables may be allowed to take every real number as a value. For a smooth function $f : \mathbb{R}^n \to \mathbb{R}^m$ its first derivative is an $m \times n$ matrix, the entries of which are the partial derivatives of the components of $f$. It is usually called the Jacobian matrix, maybe familiar to you?

However, for higher order derivatives the definition is more complicated. The second derivative of $f$ at a point $\mathbf{x}$ is a bilinear form, the third derivative a trilinear form and the $n$th derivative will be an $n$-linear form. Usually physicists call these forms "tensors", and like in the one-variable case they have an interpretation as "the derivative of the derivative of the derivative of...", evaluated at $\mathbf{x}$. When $f$ is sufficiently smooth, the multilinear forms can be expressed in terms of $n$th order partial derivatives.

I don't think it makes sense to write more about this here. I would follow the recommendations above and first become fluent in one-variable calculus (and perhaps a bit of analysis, too). After that, you will be able to make the step to multi-variable calculus much more easily. Just as a reassurance: In a typical first course on multi-variable calculus, you will not need to be concerned with multilinear forms (maybe with the exception of the "Hessian", i.e. the second derivative for the case $m = 1$), but I commented on them in view of your question. For instance, you can understand at least the statements of the implicit and inverse function theorems without being familiar with multilinear forms.

19. Nov 16, 2015

If you had bad education in UK then my education is the worst , here we don't utilize the student , they just use the "Force-Feed" method .

20. Nov 16, 2015

Thanks for your reply , That's a lot to talk about , I must be missing lots of things , I think most of the functions I have in my study their domain ∈ ℝ and some have their domain ∈ ℝ- {constant} (just one variable) we don't have anything similar to what you've said , and as I said before , I'm studying 101 calculus .

21. Nov 16, 2015

### Krylov

I know but you said:
when someone asked you about the number of variables you are interested in, so then I became motivated to write this. Don't worry you are missing things, they will come in time.

22. Nov 16, 2015

### Erland

Nader, don't you have textbook? If you don't, there are plenty of tutorials on the net which introduce derivative gently.

23. Nov 16, 2015

I have a textbook indeed , but it doesn't explain the concept clearly as it depends on how to get the derivative not what it is , for me getting the derivative of a function is pretty easy but personally I like to fully understand what I study so that's what got me here , feel free to show me any tutorial , thanks .

24. Nov 17, 2015

### Erland

Nader, I sympathize with your ambition. Too many students learn how to calculate derivatives without understanding what they are doing. So I'm happy that you want to understand.

Now, at a quick search, I don't find any tutorials which contain what I would like, so instead, let me give you some exercises, based upon the following scenario:

You drop a rock from a high tower, and it falls freely. After $t$ seconds it has fallen $y(t)=4.9 t^2$ metres.

1. Calculate the distance the rock has fallen after $3$ seconds. that is: $y(3)$.

2. Calculate the distance the rock has fallen after $4$ seconds, that is: $y(4)$.

3. Continue to calculate $y(3.1)$, $y(3.01)$, and $y(3.001)$. Write down the values in a table.

4. Calculate the distance the rock has falls between $t=3$ and $t=4$, that is $y(4)-y(3)$.

5. Continue to calculate $y(3.1)-y(3)$, that is, the distance the rock falls between $t=3$ and $t=3.1$.

6. Similarly, calculate $y(3.01)-y(3)$ and $y(3.001)-y(3)$. Write down all this in a new column in your table.

7. Calculate the average speed of the rock between $t=3$ and $t=4$, that is $\frac{y(4)-y(3)}{4-3}$.

8. Continue to calculate the average speed of the rock between $t=3$ and $t=3.1$, that is $\frac{y(3.1)-y(3)}{3.1-3}$.

9. Similarly, calculate the average speed of the rock between $t=3$ and $t=3.01$, and between $t=3$ and $t=3.001$. Write down the average speeds in yet a new column in your table.

10. Now if you look at your table, what would you say happens with the average speed over the time interval $[3,t]$ when it becomes smaller and smaller, that is, when $t$ approaches $3$ (from the right)? You may test your guess by making similar calculations for $t=3.0001$, $t=3.00001$, etc.

11. You could make a similar table to see what happends if $t$ approaches $3$ from the left instead, that is, using $t=2$, $t=2.9$, $t=2.99$, etc. instead. Does the average speed approach the same value in this case too?

12. You should by now have concluded that the average speed of the rock over a time interval which has $3$ as its left or right endpoint approaches a particular value when the interval gets smaller and smaller, that is, when the length approaches $0$ (but never becomes $0$, because then the denominator becomes $0$, so we cannot caluculate the average speed in that case). So this value is the limit of the average speed at $t=3$ as the time interval length approaches $0$.

13. This limit of the average speed at $t=3$ is precisely what we mean with the instantaneous speed of the rock at $t=3$, that is, the instantaneous speed is defined as this limit. If we could glue a working speedometer to the rock, it would show this very value at $t=3$. Let us denote this instataneous speed as $v(3)$.

14. You could make similar tables to estimate the intantaneous speed at other times, e.g. $t=1$, $t=2$, $t=4$ etc. to find $v(1)$, $v(2)$ etc.

15. If you have learned how to calculate limits, you could even calculate the instantaneous speed at a general time: $v(t)$, getting results with certainty instead of just guesses based upon tables.

16. Now, apparently, you have learned how to calculate derivatives of simple functions, so you can calulate the derivative $y'(t)$ of $y(t)=4.9t^2$.

17. Compare the derivative of $y(t))$ at $t=3$, i.e. $y'(3)$, with $v(3)$. Conclusion? If you could calculate $v(t)$ for a general $t$ at step 15, you can affirm your conclusion.

18. Some notation: at time $3$: $\Delta t = t-3$. $\Delta y= y(t)-y(3)$. Alternative notation: $h=t-3$, so $\Delta t=h$ and $\Delta y= y(3+h)-y(3)$. The derivative can be written as $v=\frac {dy}{dt}=\lim_{\Delta t\to 0}\frac{\Delta y}{\Delta t}$ (at time $3$ and at any times).

19. $v(t)=y'(t)$ is itself a function of $t$ which also has a derivative, which also can be calculated by simple rules for differentatiation (in this case) or by calculating limits, and it can be estimated by tables like these above. By definition, the derivative of $y'(t)$ is $y''(t)$. It is called the second derivative of $y(t)$. It is, by definition, the acceleration of the rock at $t$. We can continue and get the third derivative, the fourth, etc.

20. In this example, we had time ($t$) as the independent variable. Speed is the derivative (rate of change) of the distance with respect to time, and acceleration is the derivative (rate of change) of speed with respect to time. This is perhaps the most illustrative cases, but it needs not be so. Often one has $x$ as the independent variable, and we can in principle use any letter. If we blow up a baloon, we could take the derivative (rate of change) of the volume with respect to the radius. We can take the derivative (rate of change) of the tax with respect to the income, etc.

Last edited: Nov 17, 2015
25. Nov 17, 2015