1. Mar 3, 2008

### Pacopag

1. The problem statement, all variables and given/known data
Suppose you have a function $$f(x)$$.
You take the derivative with respect to x and evaluate it at some point $$x_0$$.
i.e. $${{df(x_{0})}\over{dx}}$$ (e.g. first coeff in taylor series)
Is this the same as changing the argument of f to x_0, i.e. writing $$f(x_0)$$,
then just taking the derivative with respect to x_0?
In fewer words, can I write
$${{df(x_{0})}\over{dx}}={{df(x_{0})}\over{dx_0}}$$.

2. Relevant equations

3. The attempt at a solution
If I just consider a bunch of examples, it always seems to be true. It seems like a trivial fact, but I would just like someone to confirm this for me.

Thanks.

2. Mar 3, 2008

### jhicks

No,

$$\frac{df(x_0)}{dx}$$ may be read as the derivative of f(x) with respect to x, evaluated at $$x = x_0$$. I am confused as to how you were taking the derivative of something with respect to a constant.

3. Mar 3, 2008

### Pacopag

You're right. It seems highly unorthodox.
Ok. Consider the taylor series (keeping only first order)
$$q(\Delta t)=q(0)+{{\partial H(q,p)}\over{\partial p}}\Delta t$$
$$p(\Delta t)=p(0)-{{\partial H(q,p)}\over{\partial q}}\Delta t$$
where the derivatives are evaluated at t=0. I used hamilton's equation.
$${{dq}\over{dt}}={{\partial H}\over{\partial p}}$$
$${{dp}\over{dt}}=-{{\partial H}\over{\partial q}}$$
Now regard this as a transformation from the old coords q(0), p(0) to the new coords q(Delta t), p(Delta t).
What is the Jacobian for this tranformation?