For an abrupt pn junction the relationship between the voltage and the depletion width is, $W \propto \sqrt{V_{bi}-V)}$, while for a linearly graded pn junction, $W \propto (V_{bi}-V))^{1/3}$. More generally it can be shown that if the charge density in the depletion region has the form $\rho = e\alpha \,\mathrm{sgn}(x)\mathrm{abs}(x)^m$, then the depletion width $W \propto (V_{bi}-V)^{\frac{1}{m+2}}$. For an abrupt pn-junction, $m=0$ and for a linearly graded junction, $m=1$. For $m= -1$ the depletion width is linear in $(V_{bi}-V)$ while for $m=-3/2$, $W \propto (V_{bi}-V)^{2}$.
Consider a doping profile such that the charge density has the form,
\begin{equation*} N_D(x) - N_A(x) = \alpha \mathcal{H}\Big(x+\frac{W}{2}\Big) \mathcal{H}\Big(\frac{W}{2} - x\Big) \mathrm{sgn}(x) \mathrm{abs}(x)^{m} = \frac{\rho(x)}{e} \end{equation*}with $\alpha$ being the generalized doping distribution and $\mathcal{H}$ the Heaviside function. Inside the depletion width, $-\frac{W}{2} < x < \frac{W}{2}$, the charge density $\rho$ can be simplified to
\begin{equation*} \rho = e \alpha \cdot \mathrm{sgn}(x) \cdot \mathrm{abs}(x)^{m} \end{equation*}In the following, the $\mathrm{sgn}(x)$ is omitted for simplicity and the integrals are done for $0 < x < W/2$ and are symmetric. One has to keep in mind that the space charge is negative on the p-side.
By integrating the charge density, the electric field $E$ can be determined.
\begin{equation*} E(x) = \int \dfrac{e \alpha x^m}{\epsilon} dx = \dfrac{e \alpha x^{m+1}}{\epsilon (m+1)} + C \end{equation*}Solving for the integration constant $C$ with the condition $E(x=W/2) = 0$ gives
\begin{equation*} E(x=W/2) = \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} + C = 0 \end{equation*} \begin{equation*} C = - \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} \end{equation*} \begin{equation*} E(x) = \dfrac{e \alpha x^{m+1}}{\epsilon (m+1)} - \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} \end{equation*}Integrating over the electric field gives the electrostatic potential as follows
\begin{equation*} \phi(x) = - \int E(x) dx = - \int \dfrac{e \alpha}{\epsilon (m+1)} \Big[ x^{m+1} - (W/2)^{m+1} \Big] dx \end{equation*} \begin{equation*} \phi(x) = \dfrac{e \alpha x}{\epsilon} \dfrac{\Big( (m+2) (W/2)^{m+1} - x^{m+1} \Big)}{(m+1)(m+2)} + C \end{equation*}With the conditions $\phi(x=0) = 0$ and $\phi(x=W/2) = (V_{bi} - V)/2$ the depletion width $W$ as a function of the built-in voltage $V_{bi}$ and the applied bias voltage $V$ can be determined as follows
\begin{equation*} \phi(x=W/2) - \phi(x=0) = \dfrac{e\alpha (W/2)}{\epsilon} \dfrac{\Big( (m+2) (W/2)^{m+1} - (W/2)^{m+1} \Big)}{(m+1)(m+2)} \end{equation*} \begin{equation*} \dfrac{V_{bi} - V}{2} = \dfrac{e\alpha}{\epsilon} \dfrac{(W/2)^{m+2}}{(m+2)} \end{equation*} \begin{equation*} W = 2\Bigg( \dfrac{\epsilon (m+2) (V_{bi} - V)}{2 e \alpha} \Bigg)^{\tfrac{1}{m+2}} \propto \Big(V_{bi} - V \Big)^{n} \end{equation*}with the voltage sensitivity $n = \dfrac{1}{m+2}$.
The capacitance per area of the junction $C_j$ is determined by the relation
\begin{equation*} C_j = \dfrac{\epsilon}{W} \end{equation*} \begin{equation*} C_j = \dfrac{1}{2}\Bigg( \dfrac{2 e \alpha \epsilon^{m+1}}{(m+2) (V_{bi} - V)} \Bigg)^{n} \propto \Big(V_{bi} - V\Big) ^{-n} \end{equation*}