# I Derivative with several terms in denominator

1. Feb 6, 2017

### davidge

Hi. I want to solve $\frac{\partial x^{\nu}}{\partial x^{\mu} + \xi ^{\mu}}$, knowing that $\frac{\partial x^{\nu}}{\partial x^{\mu}} = \delta ^{\nu}_{\mu}$. How can I do this?

2. Feb 6, 2017

### andrewkirk

What do $\partial x^\nu$ and $\partial x^\mu$ denote?

3. Feb 6, 2017

### davidge

$\partial$ is the symbol for partial derivative and $x^{\rho}$ is the coordinate of a point $x$.

4. Feb 6, 2017

### andrewkirk

In that case the expression in the OP has no meaning. It is simply a misuse of the partial derivative symbol.

5. Feb 6, 2017

### davidge

No. It is supposed to be a derivative. I must evaluate the derivative of $x^{\nu}$ with respect to $x^{\mu}+ \xi^{\mu}$.

6. Feb 6, 2017

### andrewkirk

In that case, there are necessary parentheses missing in the OP. It needs to be written
$$\frac{\partial x^\nu}{\partial (x^\mu+\xi^\mu)}$$
and $x^\nu$ needs to be specified as a function of $x^\mu+\xi^\mu$. What is that function? Perhaps if you provided more information about the context of your question, the function would become apparent.

7. Feb 6, 2017

8. Feb 6, 2017

### davidge

I didn't notice any two or more terms in the denominator of those derivatives.

Yes. I'm sorry.

I was trying to relate the components of a vector in the new $x$ coordinates with that in the $y$ coordinates. They should change as $$V^{\nu}(x) = \frac{\partial x^{\nu}}{\partial (y^{\mu} = x^{\mu}+ \epsilon \xi^{\mu}(x))}V'^{\mu}(y).$$

There was missing the $\epsilon$ (|$\epsilon$| << 1) in the OP, because points $y$ and $x$ are very close from each other.

I found the solution for this derivative in books of GR. It involves expanding something, where one gets terms in higher orders in $\epsilon$, there was also a minus sign. But can't remember more than this...

Last edited: Feb 6, 2017