# How do we obtain the canonical parameter from the variance function in the context of exponential family distributions?

I am assuming the following definition of the exponential family: \begin{align*} f(y,\theta,\phi) = \exp\left\{\phi[y\theta - b(\theta)] + c(y,\phi)\right\} \end{align*}

In such context, the variance function is defined as $$V(\mu) = b^{\prime\prime}(\theta)$$, where $$\mu = b^{\prime}(\theta)$$. My question is: how do we recover $$\mu$$ from $$V$$? In the textbook that I am studying, it gives the following formula \begin{align*} \theta = \int V^{-1}(\mu)\mathrm{d}\mu \end{align*}

However I am not comfortable about it. I think the author has used the informal manipulation \begin{align*} V(\mu) = \frac{\mathrm{d}\mu}{\mathrm{d}\theta} \Rightarrow \mathrm{d}\theta = \frac{\mathrm{d}\mu}{V(\mu)} \Rightarrow \theta = \int V^{-1}(\mu)\mathrm{d}\mu \end{align*}

Is it right to do so? Can someone shed some light on the problem?