I have looked at examples and forensically concluded the result is either 1 or x, when deriving wrt x.

X

How is it done in a simpler example.
(x^2+y^2+z^2) times [xihat+yjhat+zkhat divided by sqrt (x^2+y^2+z^2)]

Relevant page
<a href="https://www.quora.com/What-is-the-derivative-of-a-unit-vector">What is the derivative of a unit vector? - Quora</a>
What I've done so far
I have looked at examples and forensically concluded the result is either 1 or x, when deriving wrt x.

This is actually beyond the scope of the IntMath Forum, but we'll have a go.

The most common kind of differentiation problem involving vector quantities is wrt `t`. However, with respect to `x`, I don't believe we'll end up with a simple `1` or `x`.

Let's just do the simpler case before including the normalisation expressions. Can you differentiate this part wrt `x`?

`x hat i + y hat j + z hat k`

(BTW, click on the "Show code" button to see how to enter that expression so it comes out looking like "real" mathematics.)

X

Hello 6761
This is actually beyond the scope of the IntMath Forum, but we'll have a go.
The most common kind of differentiation problem involving vector quantities is wrt `t`. However, with respect to `x`, I don't believe we'll end up with a simple `1` or `x`.
Let's just do the simpler case before including the normalisation expressions. Can you differentiate this part wrt `x`?
`x hat i + y hat j + z hat k`
(BTW, click on the "Show code" button to see how to enter that expression so it comes out looking like "real" mathematics.)

Let's cut to the chase. What is the deriv of xi hat wrt x? Taking deriv of i,j,k is not addressed in calculus books. i,j,k are not numbers. In Cartesian coord x and xi hat are the same thing. So maybe I should just ignore the i,j,k and derive it as x,y,z.
So deriv of x,y,z wrt x is 1. Yes?

X

Let's cut to the chase. What is the deriv of xi hat wrt x? Taking deriv of i,j,k is not addressed in calculus books. i,j,k are not numbers. In Cartesian coord x and xi hat are the same thing. So maybe I should just ignore the i,j,k and derive it as x,y,z.
So deriv of x,y,z wrt x is 1. Yes?

The answer to your final question is no (in the general sense). I get the feeling you are not sure what the expression

`x hat i + y hat j + z hat k`

actually means. Can you tell me your understanding of it? (I think this is where you are having a block with this question. Let's sort that out first.)

You are right, there is no derivative for `hat i`, since it's not a variable quantity.

You are encouraged to use the math entry system so others reading this will find the mathematics easier to follow.

X

The answer to your final question is no (in the general sense). I get the feeling you are not sure what the expression
`x hat i + y hat j + z hat k`
actually means. Can you tell me your understanding of it? (I think this is where you are having a block with this question. Let's sort that out first.)
You are right, there is no derivative for `hat i`, since it's not a variable quantity.
You are encouraged to use the math entry system so others reading this will find the mathematics easier to follow.

Since i cannot find out how to do derivation of x ihat, I have come up with my own extrapolation.
d/dx (xi hat, yj hat, zk hat)=1+0+0=1
This is the only way the following derivation could be true.
A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)]
DAx
------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x)
dx
Agreed?

It seems the derivative of sin(pi*x)*ihat is the same as the sin (pi*x), further proof that the derivative of xihat wrt x is 1.
Does this seem correct?

X

It seems the derivative of sin(pi*x)*ihat is the same as the sin (pi*x), further proof that the derivative of xihat wrt x is 1.
Does this seem correct?

in Cartesian coordinates, qq i hat qq is a unit vector along the x axis, ??and x is the magnitude in that direction, and the same for j and k wrt y and z.

3 questions:

1. I'm not positive though if qq xi hat qq is a unit vector or a basis vector (or both). Could you please clarify?

2. If there is no derivative of xi,yj,zk then why have them at all in the numerator of the equation to be derived I added earlier? To me it is like saying differentiate (undifferentiable terms/sqrt(x^2+y^2+z^2) which is meaningless.

3. Why complicate matters with i,j,k when you could merely say 7x instead of 7 x i hat and they mean the exact same thing.

btw tried using the math system to type a hat but couldn't find the method.

X

Thanks.
My interpretation is it means
in Cartesian coordinates, qq i hat qq is a unit vector along the x axis, ??and x is the magnitude in that direction, and the same for j and k wrt y and z.
3 questions:
1. I'm not positive though if qq xi hat qq is a unit vector or a basis vector (or both). Could you please clarify?
2. If there is no derivative of xi,yj,zk then why have them at all in the numerator of the equation to be derived I added earlier? To me it is like saying differentiate (undifferentiable terms/sqrt(x^2+y^2+z^2) which is meaningless.
3. Why complicate matters with i,j,k when you could merely say 7x instead of 7 x i hat and they mean the exact same thing.
btw tried using the math system to type a hat but couldn't find the method.

The expression `x hat i + y hat j + z hat k` is a position vector. The `hat i`, `hat j` and `hat k` are unit vectors in the x-, y and z directions respectively.

The terms x-, y and z in the expression stand for values multiplying each of those unit vectors.

As far as the derivative is concerned, if y and z are just constants, then, yes, the derivative of the expression will be just `hat i` (or written as a coordinate vector, `(1, 0, 0).`

But let's see what happens if (say) `y` is a function of `x` (say `y = 1- x^2`, and `z` is a different function of `x` (say `z = x^3 - 2x`).

Then we would have:

`d/dx(x hat i + y hat j + z hat k)`

`= d/dx(x hat i + (1 - x^2) hat j + (x^3 - 2x) hat k)`

`= hat i - 2x hat j + (3x^2 - 2) hat k`

So I hope you can see the derivative is not just simply `1` for all cases. If however, y and z are constants, then yes, the derivative will be a unit vector in the `x`-direction, written as `hat i`.

Hope it makes sense now.

BTW, when you are logged in, you can see at the bottom of any post a "Show code" button. When you click on that, you'll be able to see how each math expression has been entered, including hat.

X

Hi again.
The expression `x hat i + y hat j + z hat k` is a <b>position</b> vector. The `hat i`, `hat j` and `hat k` are <b>unit vectors</b> in the <i>x-</i>, <i>y</i> and <i>z</i> directions respectively.
The terms <i>x-</i>, <i>y</i> and <i>z</i> in the expression stand for values multiplying each of those unit vectors.
See the 4 videos in this <a href="https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/position-vector-functions/v/position-vector-valued-functions">Khan Academy lesson</a> for more on this.
As far as the derivative is concerned, if <i>y</i> and <i>z</i> are just constants, then, yes, the derivative of the expression will be just `hat i` (or written as a coordinate vector, `(1, 0, 0).`
But let's see what happens if (say) `y` is a function of `x` (say `y = 1- x^2`, and `z` is a different function of `x` (say `z = x^3 - 2x`).
Then we would have:
`d/dx(x hat i + y hat j + z hat k)`
`= d/dx(x hat i + (1 - x^2) hat j + (x^3 - 2x) hat k)`
`= hat i - 2x hat j + (3x^2 - 2) hat k`
So I hope you can see the derivative is not just simply `1` for all cases. If however, <i>y</i> and <i>z</i> are constants, then yes, the derivative will be a unit vector in the `x`-direction, written as `hat i`.
Hope it makes sense now.
BTW, when you are logged in, you can see at the bottom of any post a "Show code" button. When you click on that, you'll be able to see how each math expression has been entered, including hat.

2 questions please...
1. But you see how after taking the derivative you ended up with i, j and k in your answer. In the derivative below there is no i,j, or k in the final answer.
A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)]
DAx
------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x)
dx
What is the difference in your example of a derivative and mine whereby you have i, j or k in your answer and I do not? If you can explain that to me you effectively have answered my question.
Could it be that the i,j,k were left out of my answer by the author of the book and it could equally be written as such?
=(x^2 ihat +y^2 jhat +z^2 khat)^1/2 + xihat(x^2 ihat +y^2 jhat+z^2 khat)^)^-1/2 (2xi hat)

2. Is A (vector)=Ax (vector)+Yz (vector) any less clear to the reader as the redundandcy of A (vector)=Ax ihat (vector)+Yz jhat (vector)?

X

2 questions please...
1. But you see how after taking the derivative you ended up with i, j and k in your answer. In the derivative below there is no i,j, or k in the final answer.
A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)]
DAx
------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x)
dx
What is the difference in your example of a derivative and mine whereby you have i, j or k in your answer and I do not? If you can explain that to me you effectively have answered my question.
Could it be that the i,j,k were left out of my answer by the author of the book and it could equally be written as such?
=(x^2 ihat +y^2 jhat +z^2 khat)^1/2 + xihat(x^2 ihat +y^2 jhat+z^2 khat)^)^-1/2 (2xi hat)
2. Is A (vector)=Ax (vector)+Yz (vector) any less clear to the reader as the redundandcy of A (vector)=Ax ihat (vector)+Yz jhat (vector)?

I think I need to see the original context to work out what the notation actually means. (You can upload an image in this forum). It's not clear to me what Ax refers to.

My derivative above is for the expression `x hat i + y hat j + z hat k`, which I said we would do first and talk about the normalized case later.

The length of the vector `x hat i + y hat j + z hat k` is `sqrt(x^2+y^2+z^2)`.

So the normalized vector is:

`(x hat i + y hat j + z hat k)/sqrt(x^2+y^2+z^2)` `= (sqrt(x^2+y^2+z^2)(x hat i + y hat j + z hat k))/(x^2+y^2+z^2)` `= (x hat i + y hat j + z hat k)xx(x^2+y^2+z^2)^(-1/2)`

I'm guessing the text book's author is using `A_x` to mean "the `x`-component of the normalized vector `A`". So that would mean

This is not the answer you put (which is hard to read - please use the math entry system), but hopefully gives you a start on following what the author is doing.

X

I think I need to see the original context to work out what the notation actually means. (You can upload an image in this forum). It's not clear to me what Ax refers to.
My derivative above is for the expression `x hat i + y hat j + z hat k`, which I said we would do first and talk about the normalized case later.
The length of the vector `x hat i + y hat j + z hat k` is `sqrt(x^2+y^2+z^2)`.
So the normalized vector is:
`(x hat i + y hat j + z hat k)/sqrt(x^2+y^2+z^2)` `= (sqrt(x^2+y^2+z^2)(x hat i + y hat j + z hat k))/(x^2+y^2+z^2)` `= (x hat i + y hat j + z hat k)xx(x^2+y^2+z^2)^(-1/2)`
I'm guessing the text book's author is using `A_x` to mean "the `x`-component of the normalized vector `A`". So that would mean
`A_x = (x)xx(x^2+y^2+z^2)^(-1/2)`
In which case,
`(dA_x)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x + 2y(dy/dx + 2z dz/dx))` ` + (x^2+y^2+z^2)^(-1/2)(1)`
If `y` and `z` are constants, then this would give us:
`(dA_x)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x)` ` + (x^2+y^2+z^2)^(-1/2)`
`= -x^2(x^2+y^2+z^2)^(-3/2)` ` + (x^2+y^2+z^2)^(-1/2)`
This is not the answer you put (which is hard to read - please use the math entry system), but hopefully gives you a start on following what the author is doing.

Attached is the context regarding divergence of electric fields.
The author does get a different result than you do, although I've found errors in his text several times.

Can you tell me why i,j,k are not eliminated when you executed your derivative but is eliminated in the text book example I attached?
Also, if after reading the context you now agree with the textbook, can you please show me all the steps in the derivation in extreme detail, since I cannot understand how the book came up with its result. Thank you!

X

Thanks.
Attached is the context regarding divergence of electric fields.
The author does get a different result than you do, although I've found errors in his text several times.
Can you tell me why i,j,k are not eliminated when you executed your derivative but is eliminated in the text book example I attached?
Also, if after reading the context you now agree with the textbook, can you please show me all the steps in the derivation in extreme detail, since I cannot understand how the book came up with its result. Thank you!