How Do You Differentiate a Vector Function with Respect to Another Vector?

[Sorento7]

Homework Statement

\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}

b is not a function of a

Homework Equations

I want to differentiate this, (the jacobian of the vector field)
dot is the Euclidean inner product.

The Attempt at a Solution

\acute{u}.v - \acute{v}.u / v² doesn't seem to work

 

[DryRun]

To begin with, i could barely see anything with that font size...
\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}

 

[Sorento7]

\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}

 

[Sorento7]

no answer? wow I thought that should have been an easy differentiation?

 

[I like Serena]

It's not a proper differentiation.
I can't make out what your formula is supposed to represent.
It doesn't look like a Jacobian.

 

[Sorento7]

Maybe I phrased it wrong. Its the result of multiplying inversion of an inner product to one of its vector components. (1/\vec{a}.\vec{b}) * \vec{a}

edit: \vec{b} is a constant vector.

 

[I like Serena]

So is $\vec a$ a function of (x,y) or something?

 

[Sorento7]

So is $\vec a$ a function of (x,y) or something?

Suppose:

F(\vec{a}) = (1/\vec{a}.\vec{b}) \vec{a}

dot is the Euclidean inner product and F is defined as a vector space(R³ → R³)

I need ∂\vec{F}/∂\vec{a} (given that \vec{b} is an arbitrary constant vector.)

 

[I like Serena]

Ok, so suppose $\vec a$ is $[^x_y]$.

Then $\vec F(\vec a) = \vec F(x, y) = [^{F_x(x,y)}_{F_y(x,y)}]$.

In that case the Jacobian is the 2x2 matrix of the partial derivatives of F.

With your $\vec F(\vec a) = {\vec a \over \vec a \cdot \vec b}$, you get:
$$\vec F(\vec a) = {[^x_y] \over x b_x + y b_y}$$
From this you can calculate the partial derivatives.

For instance:
$${\partial F_x \over \partial x} = {\partial\over \partial x}({x \over x b_x + y b_y})$$
With an application of the quotient rule the result follows...

 

[Sorento7]

Thanks. That was helpful