 Solution

Differentiate the normalization of the exponential family density with respect to ${\theta }_i$, assuming the regularity that permits differentiation under its integral. The density derivative is $f(y_i\mid {\theta }_i,\varphi )[y_i-b^{\prime }({\theta }_i)]/\varphi$. ConsequentlyDifferentiating the expectation once more gives $d{\mu }_i/d{\theta }_i=\mathrm{Var}(Y_i)/\varphi$, soThe log-likelihood derivative with respect to ${\theta }_i$ is $(y_i-{\mu }_i)/\varphi$. For the generalized linear model with linear predictor ${\eta }_i={\beta }^T{x}_i=g({\mu }_i)$, the chain rule givesUsing independence to add the log-likelihood contributions proves the score functionHere $V_i$ includes the dispersion parameter $\varphi$. Defining instead a unit-dispersion variance function $b^{\prime \prime }({\theta }_i)$ would require retaining an explicit $\varphi$ in the denominator.