Find the expected value of "Y", exponential family with lots of questions here I have a problem I don't know how to approach. It is

A generalization of the 1-parameter exponential family, to allow 2-parameter distribution, is the family given by
$$f(y;\theta, \phi)=e^{\frac{y \theta - c(\theta)}{a(\phi)}+d(y,\phi)}$$
Show that $E(Y)=c'(\theta)$ and $var(Y)=c''(\theta) a(\phi)$

Well, first off, maybe this is common understanding for experienced people but is this exponential distribution a continuous one? A Discrete one?
I ask this because my only idea to approach the question is by first principles, so for the expected value of Y, I thought either integrating or summing $yf(y)$ would do (if that is possible at all) but from the information given, I don't know which to go for.
Is there a simpler way? I am confused, it would be great if someone would tell me how to solve this.
 A: Hint. (not meant to be a complete solution)
Exponential families are very different from the usual exponential distribution (but of course, the exponential distribution is a special case of a distribution in the exponential family). Distributions that are of an exponential family can be either continuous or discrete. Here, I'm going to prove the claim for the continuous case.
Suppose $Y$ is continuous, and is of the two-parameter exponential family given above.
Without getting into technical details, assume that interchange of differentiation and integration is allowed. We know that $$\int_{\mathbb{R}}f(y; \theta, \phi)\text{ d}y = 1\text{.}$$
Differentiate both sides to get
$$\dfrac{\partial}{\partial \theta}\int_{\mathbb{R}}f(y; \theta, \phi)\text{ d}y = \int_{\mathbb{R}}\dfrac{\partial}{\partial \theta}f(y; \theta, \phi)\text{ d}y = 0\text{.}\tag{1}$$
Now
$$\dfrac{\partial}{\partial \theta}f(y; \theta, \phi) = e^{d(y, \phi)}\exp\left[\dfrac{y\theta-c(\theta)}{a(\phi)} \right](y-c^{\prime}(\theta)) = \frac{y-c^{\prime}(\theta)}{a(\phi)}f(y;\theta, \phi)$$
so that $(1)$ implies that
$$\int_{\mathbb{R}}[y-c^{\prime}(\theta)]f(y;\theta, \phi)\text{ d}y = \int_{\mathbb{R}}yf(y;\theta, \phi)\text{ d}y - \int_{\mathbb{R}}c^{\prime}(\theta)f(y;\theta, \phi)\text{ d}y = 0\text{.}\tag{2}$$
Notice two things: that $c^{\prime}(\theta)$ is not dependent on $y$, and do you recognize what
$$\int_{\mathbb{R}}yf(y;\theta, \phi)\text{ d}y$$
is?
For a second run, it is shown similarly that$$\int_{\mathbb{R}}\dfrac{\partial^2}{\partial \theta^2}f(y; \theta, \phi)\text{ d}y = 0\text{.}\tag{3}$$
and I leave it as an exercise to you to show that
$$\begin{align}
\dfrac{\partial^2}{\partial \theta^2}f(y; \theta, \phi) &= \left(\frac{y-c^{\prime}(\theta)}{a(\phi)}\right)^2 f(y;\theta, \phi) - \frac{c^{\prime\prime}(\theta)}{a(\phi)}f(y;\theta, \phi) \\
&= \frac{y^2 - 2yc^{\prime}(\theta) + [c^{\prime}(\theta)]^2}{[a(\phi)]^2}f(y; \theta, \phi)- \frac{c^{\prime\prime}(\theta)}{a(\phi)}f(y;\theta, \phi)\text{.}
\end{align}$$
Integrate this over $\mathbb{R}$, set it equal to $0$ (i.e., do $(3)$), and proceed similarly to how $(2)$ is derived. Furthermore, use the result of $(2)$ to do $(3)$. Be sure to simplify this expression. This will finish your proof for the continuous case. The summation case is similar.
