Transformation of differential operators by blowing up M is a smooth space, $\sigma:M'\rightarrow M$ the blowing-up, centre is a smooth closed subspace C. A local coordinate chart U of M with coordinates $(x_1,...,x_n)$ in which $C=\{x_r=...=x_n=0\}$. Then $\sigma^{-1}(C)$ is covered by coordinate charts $U_{x_r},...,U_{x_n}$ and we can choose coordiantes $(y_1,...,y_n)$ with $y_1=x_1,...,y_r=x_r,y_{r+1}=x_{r+1}/x_r,...,y_n=x_n/x_r$.
Show that with coordinates chosen as above
$\frac{\partial}{\partial x_j}=\frac{\partial}{\partial y_j}$ for $j=1,...,r-1$
$x_r\frac{\partial}{\partial x_r}=y_r\frac{\partial}{\partial y_r}-\sum_{j=r+1}^n y_j\frac{\partial}{\partial y_j}$
$x_j\frac{\partial}{\partial x_j}=y_j\frac{\partial}{\partial y_j}$ for $j=r+1,...,n$
We apply the left hand side to a function $f(x)$ and then composed with $\sigma$ and the right hand side to $f\circ\sigma$.
I am revising a lecture in algebraic geometry rn and I am completely stuck. What do I need apart from the chain-rule?
 A: Assume that $U\subset M$ is given with coordinates $(x_1,\ldots,x_n)$ and that the centre of the blow-up $C$ is defined by $x_r=x_{r+1}=\ldots=x_n=0$. Then the blow-up of $U$ at $C$ can be covered by charts $U_{x_r},\ldots, U_{x_n}$ so that the coordinates of, say for example $U_{x_r}$, are $(x_1,\ldots,x_{r-1},x_r, \frac{x_{r+1}}{x_i},\dots, \frac{x_n}{x_r})=:(y_1,\ldots,y_n)$. 
Note that $\frac{\partial}{\partial x_i}$ is a derivation of the coordinate ring $A(U)$ of $U$, but it can be also exteneded to a derivation of the fraction field of $A(U)$ in the usual way (you know how to derivate a quotient of two functions). Note also that $A(U_{x_r})$ is a subring of the fraction field of $A(U)$. You can now check that $\frac{\partial}{\partial x_j}(A(U_{x_r}))\subset A(U_{x_r})$ if $j=1,\ldots,r-1$ but this is not longer true if $j\geq r$. In fact, for $j\geq r$ one has $x_r\frac{\partial}{\partial x_j}(A(U_{x_r})\subset A(U_{x_r})$ and hence $x_j\frac{\partial}{\partial x_j}(A(U_{x_r})\subset A(U_{x_r})$ since $x_j=y_r x_r$.
The conclusion is that $\frac{\partial}{\partial x_1},\ldots, \frac{\partial}{\partial x_{r-1}}, x_r\frac{\partial}{\partial x_r},\ldots,x_n\frac{\partial}{\partial x_n} $ are all derivations of $A(U_{x_r})$. What you have to prove are some equalities of derivations of $A(U_{x_r})$ and to do that you only need to evaluate these derivations at the coordinate functions of $U_{x_r}$, namely, $y_1,\ldots,y_n$. 
For example, for the first equality you have to prove that $\frac{\partial}{\partial x_j}(y_k)=\frac{\partial}{\partial y_j}(y_k)$ for $j=1,\ldots,r-1$ and all $k=1,\ldots,n$. On the right hand side you have $\delta_{jk}$. On the left hand side, we also get this result if $k\leq r$ since in this case $y_k=x_k$. If $k> r$ then $y_k=\frac{x_k}{x_r}$ and we obtain $\frac{\partial}{\partial x_j}(\frac{x_k}{x_r})=0=\delta_{jk}$ since $j< r$. 
