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All right.
I'll admit this question, this problem here that I'm going to
show you is not an easy one, all right?
It stumps a lot of students.
And so let me work this one out for you.
Kind of take it step by step.
We're given the function h of x, which is x times the
cotangent of 8 square root x.
Now this here, this 8 square root x is the argument for
cotangent, OK?
This is the angle, if you will.
This is the angle for cotangent.
And then plus 7, some constant hanging out here at the end.
Before I get too deep with this and find the derivative
of it, let me show you, because we're going to need
this when we use the chain rule in just a moment, let me
show you how do you find the derivative of 8 square root x?
All right, so maybe that's something we can work out real
quickly here.
Let's go find the derivative of 8 square root x.
Move this out of the way for a second.
OK, so if we have 8 square root x, which if it's OK with
you, I'm going to write this way.
8x to the 1/2, right?
Because a square root is really just a fractional
exponent of 1/2.
Well, you know that the derivative rule says OK, well
let's take that 1/2 and multiply it by whatever's out
front here.
And then we'll subtract 1 from this, right?
We'll subtract 1 from 1/2.
But I'm going to write the 1 this way.
I'm going to write it as 2 over 2.
And 1/2 minus 2 over 2 is simply just a
negative 1/2, OK?
So I have x with a negative 1/2 now.
But look what number I have out front.
Between this 1/2 times 8 is really just a 4.
OK, so I have 4x to the negative 1/2.
And since I don't really like negative exponents too much,
I'm going to bring that down and write this as 4 over x to
the positive 1/2.
And again, if you don't like fractional exponents, you
could even go one step further and write it
this way if you want.
OK?
But all three of these, I hope you see all three of these
forms right here, I'll circle them for you, are equivalent
to each other.
This form here, that form there, and that form there.
All three of those forms are exactly the same.
So let's keep that in mind.
If you want to find the derivative of 8 square root x,
you can write it in any one of those two forms.
Four's in the numerator, square root
of x is on the bottom.
OK, now armed with that information, let me set that
aside because we might need that later.
Let's go off and find the derivative of this function
here, all right?
Of this function here.
I'm going to use the product rule on this, and then you're
going to see in just a bit, I'm going to need the chain
rule as well.
Product rule says since, by the way, the derivative of 7
is just a 0 anyway, so I'm not really going to even bother
with the 7.
I'm just going to ignore it completely for now, OK?
So let's call this part of the product here our f, and let's
call this part over here our g.
OK, so product rule says, let me scoot this over.
Product rule says first, give me the
derivative of f, all right?
Right so here is my, maybe I should do it this way.
Put h prime over here, right?
So the derivative of the x is simply just a 1 times the
derivative of g.
Or times, rather, times just g as it is.
I'm not even taking the derivative of it yet.
So I'm just going to leave it like so.
And that's not going to change any, right?
Because 1 times cotangent of stuff is just
cotangent of stuff.
Plus, here's the rest of the product rule.
Plus my f, which in this case is x times, all right, I'll
put that in parentheses.
Times the g prime.
Times the derivative of all of this.
And now here's where I'm going to need the chain rule, OK?
The chain rule.
And that's because it's not just cotangent of x.
But there is inside here, this argument for cotangent is
another function in and of itself.
So let's do this first.
How about we do this?
I'm going to cover up this stuff inside, and just ask
what is the derivative of cotangent?
What's the derivative of cotangent?
Oh, the derivative of cotangent is negative cosecant
squared of whatever that stuff was, right?
Whatever that stuff inside was, which was 8
square root x, OK?
But the chain rule says go one more step.
Now, give me the derivative of the inside, this inner part.
Well, remember we just did that a second ago?
The derivative of that inner stuff, right, was 4x to the
negative 1/2.
So I need to squeeze that over here as well.
Am I going to have room?
4x to the negative 1/2, barely have room for that, OK?
So here's what we've got.
This is really huge and nasty, and again, the derivative, if
I kept going, the derivative of this
constant is just a , anyway.
So I'm not even going to bother with that.
So let me clean this up a little bit more, because we
can simplify that down.
That really is our answer, but let's simplify it down a bit.
OK, so I've got prime is equal to, and again, one times this
cotangent is not going to change it any, so I'm just
going to simply write it as cotangent of 8 square root x.
And now I need to clean up this product over here.
I've got three things that I'm multiplying together.
Three things.
First of all, I see a negative sign right there, so how we
move the negative out all the way?
Because a positive times that negative is just
going to be a negative.
I also see a coefficient of 4, so I might as well move that
all the way out as well.
And here's a couple of things going.
On do you see this x times this x to the negative 1/2?
I kind of almost covered up the
negative sign there, right?
So I have this x times this x to the negative 1/2.
Now, maybe, let me bring this back for a second.
What is the product of x and x to the negative 1/2?
Well, you know that we simply, all right, if I write this as
x to the first power, we simply add exponents here,
since the bases are the same.
But I don't like writing 1 this way, since this one
already has a denominator of 2.
So how about if I write this 1, this exponent of 1 like
this, 2 over 2.
I could write it that way if I wanted to, right?
And now if you see, if you add these, you get a
positive 1/2, right?
The product of x to the first power times x to the negative
1/2 is simply x to the positive 1/2.
Or you could write it as the square root of
x, same thing, right?
So let's write it this way.
Let's write the product of these two things right here,
these two bases of x as the square root of x.
Cool.
I like that.
OK, and then here's the rest of it.
The rest of it is this cosecant squared.
And its argument of 8 square root of x.
So there is my final answer.
You could write it like this, or you could put this term
first, and then you put the cotangent.
And you could write it either way you want.
Doesn't really matter.
How about if I write it both ways for you?
You could write it this way if you wanted to.
Cosecant squared.
It's argument of 8, square root of x.
Plus, and then this term over here, this cotangent.
Here we go.
Cotangent of 8, square root of x.
So either one of those two are the correct answers.
I kind of like this one, I'm partial to this one since it's
the first one I came up with.
And I hope that makes sense to you.
So we needed a product rule as well as inside the product
rule, we needed the chain rule.
Hope that helps.