# Calculus I - Lesson 3: Properties of the derivative

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Hello again!

I know it's been rather a while since the last thread, but I'm hoping that someone's still out there I know there's been a fair amount of demand for this to continue, so I plan on posting another few at least. I really must apologise for my absense; things have been happening rather fast around here recently!

So, this lesson is all about properties of the derivative and what you can do with it. I feel this may blend a little into the next topic I planned to cover, but nevermind. The answers from last time's post are all in the previous thread - if they're not, just ask me for them.

Some properties

Right. So, we've gone off and differentiated our nice little function. Now what?

Well, there's a few issues that we need to bring up. First of all, I don't think I really explained myself very well last time when we considered some functions with constants in them; for example we know how to differentiate xn, but we don't really know how to differentiate axn - I didn't really mention this, I just assumed it which was quite bad. Let us assume that we have a function, y = axn, where a is a constant (like 2, 3.4, sqrt(2), etc). Then:

$\frac{dy}{dx} = a \frac{d}{dx}(x^n)$.

In fact, this rule applies to any function that we can think of - we certainly don't know how to find the derivative of the sine function, for example, but it's certainly true that:

$\frac{d}{dx}(a\sin(x)) = a \frac{d}{dx}(\sin(x))$.

So, now I've cleared that out of the way, some other facts about the derivative. Suppose you've got a nice quadratic curve, y = x2 + 5x + 6, and you want to find the gradient at a certain point. Our rule will only cover you for functions by themselves - what do we do about functions that are added together? Well, we can use a really simple fact about the derivative of this function:

$\frac{dy}{dx} = \frac{d}{dx}(x^2+5x+6) = \frac{d}{dx}(x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(6) = 2x + 5$.

We can just split up the derivative into sums of little bits that we know how to differentiate.

Sidenote for the interested reader: Try to prove this using first principles - it's not all that hard!

Please, please, please note that this is NOT true for something like y = x * x2. We can even show this isn't true. We know that by simplifying this down using the laws of indices, y = x3, which has the derivative 3x2. But if we apply the same kind of rules as we did before, we get the derivative being 2x instead! There is a correct way of doing this (called the product rule), and we shall cover it later.

Maxima and Minima

Suppose, again, that we have our quadratic graph y = x2 + 5x + 6. It's quite easy to find the roots are, and then go off and plot it nicely on a piece of paper. If you look at your graph, you'd notice that it dips at the bottom, and then comes back up again. We call this a critical point or turning point - a point at which the graph "turns" - just before the critical point, it was going down, and just after it's heading upwards. We call this type of critical point a minima for obvious reasons.

There is, as you would guess, another type of turning point. For example, if you reflected the graph in the x-axis, you would find a point on the graph where just before this point, the graph would be heading up and just after heading down. We call this (quite unimaginatively) a maxima. I give some examples of both of these types of points in the image at the end - they're quite easy to see, it's just a bit hard to describe them using words

Now let me give you another quadratic: y = 673x2 + 2348x - 28734. Suppose I want you to find the critical point of this. It's not easy this time - plotting the graph might be a little tricky! So, how do we do this? Well, the easiest way is to use our good old friend, the derivative.

"But how?!?!", you say. Well, let's consider what happens around a turning point. As we get closer and closer to it, you might see that the gradient of the curve gets very small. Indeed, you might even guess that at the turning point, the gradient would be zero - and you've be right.

(If you can't see this, draw any old turning point and then draw tangents as you get closer to it, then look at the gradient of the tangents).

So, if we have some function y, then we can find out where the turning points are, simply by using the fact that at a turning point, the derivative is zero. Let's try and do this for the first example and you can draw and confirm the result. We showed earlier that:

$\frac{dy}{dx} = \frac{d}{dx}(x^2+5x+6) = \frac{d}{dx}(x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(6) = 2x + 5$.

Now, we want to find out the value of x for which we have a turning point. So:

$\frac{dy}{dx} = 0 \Rightarrow 2x+5 = 0 \Rightarrow x = -\tfrac{5}{2}$.

(Remember, $\Rightarrow$ means "implies that".)

Well, that was a bit easy! We can even work out what the value of y is at this point, because we have a formula for that. (I'll leave this as an exercise).

However, the more astute amongst you may have noticed that all we've done is actually found out that we have a turning point at x = -5/2. We didn't actually find out which kind of turning point we get. To do this, we have to use something called the second derivative. Although it sounds quite technical, it isn't. When we differentiate a function, we call this the first derivative. Logic prevails, and if we differentiate again, we get the second derivative. (Indeed, if we keep doing it, we get the third, fourth, fifth, etc derivatives). We also introduce some special notation:

$\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

As you can see, it's a much shorter than writing the gubbins on the right In practical terms, it doesn't mean a lot. For example, if we have the function y = x3, then:

$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3x^2) = 6x$.

There are some more exercises to do at the bottom, but the idea behind the second derivative is fairly straight-forward. How does the second derivative relate to critical points? The simple answer is this:

If $x^*$ is a critical point and $\frac{d^2 y}{dx^2} < 0$ at $x^*$, then $x^*$ is a maxima. If $\frac{d^2 y}{dx^2} > 0$ at $x^*$, then $x^*$ is a minima.

I hope you take that all in: by $x^*$, I just mean a single point - it's just an easy bit of notation. I won't try and explain it for now; just accept it, and I will hopefully try and explain towards the end (when I do the roundup). If you didn't get it, go over and try again. It'll get there eventually

Some people may have spotted that I haven't stated what happens when $\frac{d^2 y}{dx^2} = 0$. Well, there is a third type of critical point, namely a point of inflexion - for an example, look at x3 when x = 0. You see that the gradient at x = 0 is zero, but the point is not a maxima or a minima. Points of inflexion are harder to detect and it isn't the case that if the second derivative is equal to zero, then it's a point of inflexion - other odd things can happen as well. We shan't cover that here, but it involves looking at the third derivative of the function.

So, an example. Let's take our classic quadratic, y = x2 + 5x + 6. We already showed that it has a turning point at x = -5/2. Let's try and see whether we can prove it's a minima. For this quadratic, we get:

$\frac{d^2 y}{dx^2} = 2$.

This is a particularly easy case; 2 > 0 (obviously), so we have a minima. Done!

For a more non-trivial example, let's look at a cubic equation, namely y = x3 - x. We want to find out where the critical points are on this graph and what type they are. First, let's differentiate the function:

$\frac{dy}{dx} = 3x^2 - 1$.

Now, set this equal to zero:

$\frac{dy}{dx} = 0 \Rightarrow x^2 = \frac{1}{3}$.

So now we square root both sides to obtain:

$x = \pm \frac{1}{\sqrt{3}}$.

Notice that I leave everything as it is; no decimals around here! So, we know our critical points. What type are they? To work this out, we need to find the second derivative.

$\frac{d^2 y}{dx^2} = 6x$.

Now, put our values of x in: so, when $x = +1/\sqrt{3}$, the second derivative will be $6/\sqrt{3} > 0$, so we have a minima. When $x = -1/sqrt{3}$, the second derivative will be $-6/sqrt{3} < 0$, so we have a maxima. If you plot the graph (this isn't too hard either), you'll find that this looks about right.

We would never have been able to work that out precisely by just looking at the graph either! That just about sums it up for this lesson; again, it's a lot to take in, but I plan on making some more exercises for you all tomorrow at some point. Until next time,

- Dave

Exercises

1) Calculate the derivative of the following functions:

i) $x^2+5x$

ii) $x^{-2}+17x$

iii) $\frac{3}{\sqrt{6x}} + 4x^{8/3}$

iv) $\frac{1}{x}\cdot \frac{x^2}{a^2} + a^4$, where a is constant.

2) Calculate the second derivatives of all the functions above.

3) Calculate the value of x for which the horrible polynomial for earlier will have a turning point. Give the y value as well. (You might want to use a calculator )

4) (Harder) Let's assume that we have a cylinder that we want to fill with precisely 1 litre of water (i.e. the volume of the cylinder is equal to 1 litre). Obviously, we can do this by letting the cylinder be different heights and radii - what we would like to do is to minimize the surface area of the cylinder, just for the heck of it.

a) Calculate the surface area S of the cylinder in terms of its height h, radius r and pi.

b) Using the fact that the volume is 1, express S in terms of just r (hint: use substitution).

c) Now we have a regular function, just like y=x. This time, however, S is some function of r. Calculate the turning points for the function and find a minima for S.

Question 4 is, of course, quite hard for people who have just picked all of this up. We're going to do a couple examples next time, but if you have a bit of free time, take a look!

Please, as always, rate the thread and your feedback is always appreciated.

Cheers.

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here is what i have for the first three. haven't started the fourth yet.

1)i)2x+5

ii)-2x+17

iii)$\frac{32}{3}x^{\frac{5}{3}}-\sqrt{6}x^{-3}$

iv)$a^{-2}$

2)i)5

ii)17

iii)$\frac{160}{9}x^{\frac{2}{3}}+3\sqrt{6}^{-4}$

iv)0

3)i) $(-\frac{5}{2},-6.25)$

ii)$(\frac{17}{2}, 144.514)$

iii)(.001,37.926)

iv)$(a^{-2},a^{-6}+a^4)$

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You might want to take a look at 1) iii). I'm not sure why you calculated maxima and minima for all of those, I only wanted the one from the nasty polynomial

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what nasty polynomial?

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y = 673x2 + 2348x - 28734.

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is it(-1.744,-3.958) and it is a minima?

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Ah yes, so I did. No matter, it's there now

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i thought in the other lesson you said $\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

edit: how do you get the "h approaches zero" under "lim"?

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edit: how do you get the "h approaches zero" under "lim"?

You need to use \lim instead of just lim

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is the answer to #4 $(\pi^{-\frac{1}{3}},3\pi^{\frac{1}{3}})$? If so, it wasn't that hard.

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Just general info:

A point of inflection is a point where a curve crosses its tangent. So a point of inflection doesnt have to be a stationary point. But a stationary point which is neither a maximum or a minimum has to be by default a point of inflection.

For the case where the second derivative is equal to 0 at the stationary point, one of the ways forward is to look at the power series (taylor series) of the function and how it behaves close to the stationary point.

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bloodhound, thanx for confusing me.

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hi I have a problem in washer method

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What exactly is this washer method?

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i thought in the other lesson you said $\frac{dy}{dx}=\lim_{h\to0}\frac{f(x-h)+f(x)}{h}$

edit: how do you get the "h approaches zero" under "lim"?

Dave' date=' I have a question, isn't the definition of derivative supposed to be:

[math']\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h) - f(x)}{h}[/math]

Why does yourdadonapogos have you using a different definition of derivative, than the standard one? Something appears to be going on with the direction from which h is approaching zero.

Thank you

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I don't know. My definition of the derivative appears to be the same in both threads (although the latex is garbled in one of them).

I'm also going to knock the size of these latex images down, because they're huge

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I don't know. My definition of the derivative appears to be the same in both threads (although the latex is garbled in one of them).

Can you please direct me to the other thread, where you introduce the definition of derivative.

Thank you

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Sure, just look at the Calculus I - Lesson 1/2 threads earlier in this forum.

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Sure, just look at the Calculus I - Lesson 1/2 threads earlier in this forum.

I looked through the thread entitled, "Calculus I- Lesson 1" and I couldn't find it. Which post?

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it is in lesson 2

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it is in lesson 2

A moment ago, lesson 2 read lesson 1, i think there is a problem with the server. Whatever...

Yes I looked through lesson 2, i didn't see it there, which post?

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The very first one.

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The very first one.

Ok, I saw it.

You introduce the definition of derivative in Calculus I-lesson 2, post number one.

The latex was garbled (just as you said), which was why I missed it the first time.

Now, in CalcI,lesson2,post1, you give the following as the definition of derivative:

$\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$

Now look at my post #15, where I quote yourdadonapogos, and you see that he has you saying this:

Originally Posted by yourdadonapogos

i thought in the other lesson you said

$\frac{dy}{dx}=\lim_{h\to0}\frac{f(x-h)+f(x)}{h}$

edit: how do you get the "h approaches zero" under "lim"?

I just copied that now.

Now his post is this

i thought in the other lesson you said $\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

edit: how do you get the "h approaches zero" under "lim"?

Apparently you guys are having some problems with the computer over there.

I noticed problems before, but I didn't say anything.

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i was typing fast. fixed.

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i was typing fast. fixed.

Oh ok, now it makes sense.

Regards

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