How to calculate the functional derivative of the functional integral? I study by myself with the QFT, in the page 197 of book of Lewis H. Ryder (2nd edition), The author wrote that he take the functional derivative of equation 6.69:
$$\frac {\delta\widehat{Z}[\phi]}{\delta\phi}$$
where $$\widehat {Z}[\phi]=\frac{{e}^{iS}}{\int{{e}^{iS}}{\cal D}\phi}\tag{6.69}$$  
and
$$S=-\int{\left[\frac {1}{2} \phi(\Box+{m }^{ 2 })\phi -{\cal L  }_{ int } \right] { d }^{ 4 }x }.\tag{6.71} $$
The result in Eq. 6.72 is:
$$\frac { \delta  }{ \delta \phi  } \left\{ \exp\left[ -i\int { \left[ \frac { 1 }{ 2 } \phi (\Box +{ m }^{ 2 })\phi-{\cal L}_{ int } \right]  } { d }^{ 4 }x \right]  \right\} { \left[ \int { \exp\left[ iS \right]  } {\cal D}\phi  \right]  }^{ -1 }\\=
\left( \Box +{ m }^{ 2 } \right) \phi \widehat { Z } [\phi ]-\frac { \partial { \cal L }_{ int } }{ \partial \phi  }\widehat { Z } [\phi ].\tag{6.72} $$
I don't understand how the calculating procedure taking place. I have known how to calculate the functional derivative to a functional, but I do not know how to take it to a functional integral like $\widehat{Z}[\phi]$. I would be most thankful if anyone help me. 
PS: Is there are some detailed textbook or literature about this technique?
 A: Hints to the question (v2):


*

*First of all, one should realize the abuse of notation in eq. (6.69) of Ref. 1 where $\phi$ is used in two meanings: both as an external parameter and as internal integration/dummy variable. It is more properly written as
$$ \widehat {Z}[\phi]~=~\frac{{e}^{iS[\phi]}}{{\cal N}}, \qquad {\cal N}~:=~\int\!{\cal D}\phi~e^{iS[\phi]} .\tag{6.69'}$$

*Similarly, eq. (6.72) is essentially equivalent to 
$$ \frac { \delta \widehat { Z } [\phi ] }{ \delta \phi(x)} ~=~\frac{1}{\cal N} \frac {\delta {e}^{iS[\phi]} }{ \delta \phi(x)} 
 ~=~\frac{{e}^{iS[\phi]}}{{\cal N}}\frac{\delta S[\phi ]}{\delta\phi(x) }
~=~\widehat { Z } [\phi ]\frac{\delta S[\phi ]}{\delta\phi(x) }.\tag{6.72'} $$
In the first equality we stress that we shouldn't differentiate the denominator ${\cal N}$ wrt. $\phi(x)$ because the denominator ${\cal N}$ doesn't depend on $\phi(x)$, cf. comment by David Vercauteren.

*Finally, the functional derivative $\frac{\delta S[\phi ]}{\delta\phi(x) }$ becomes the Euler-Lagrange expression.
References:


*

*L.H. Ryder, QFT; Section 6.4, p. 197.

