Mass counter term is a derivative at tree level?

[CAF123]

I've heard the statement that by computing just the leading-order (tree level) diagrams of a process and then computing the derivative of this result with respect to the mass should correspond to the evaluation of the mass counter term diagrams. Can someone explain why this statement is precisely true?

If we renormalise $m_0 = Z_m m$ then the bare amplitude $$A(m_0) \rightarrow A(Z_m m) = A(m) + m(Z_m-1) \frac{d A(m)}{dm} +O ((1-Z_m)^2)$$

But why is dA(m)/dm equal to the derivative of the tree level result?

 

[vanhees71]

It's just the Taylor expansion of
$$f(Z_m)=A(Z_m m)$$
around $Z_m=1$:
$$f(Z_m)=f(1)+(Z_m-1) f'(1)+\frac{1}{2} (Z_m-1)^2 f''(1)+\cdots$$
Now obviously
$$f^{(n)}(1)=A^{(n)}(m).$$

 

[CAF123]

@vanhees71 thanks. Why is $f^{(1)}(1) = A^{(1)}(m)$, the derivative of the tree level result, equal to the mass counter term at one loop though?