## 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

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

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

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

I like Serena
Homework Helper
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.

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

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
Homework Helper
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...