Partials of PDF with no closed form solution I need to estimate partial derivatives for all N parameters denoted $\theta_{N}$ of a probability density function(PDF) $\mathcal{f}$.  
This PDF $\mathcal{f}$ has no closed form solution and is instead expressed in terms of a characteristic function $\phi$ and an FFT $\mathcal{F}^{-1}$.
$\mathcal{f}(x) = \mathcal{F}^{-1} [\phi(x)]$
Is it possible to estimate $\frac {\partial \mathcal{f}} {\theta_{N}}$  for every N?
 A: Given your description of your problem, I'm going to assume that your characteristic function is in a closed form that is simple to differentiate, but your density isn't.  I'm also going to assume that you are capable of taking the inverse Fourier transform of an object, either analytically or by numerical methods.  To facilitate our analysis, let's denote the partial derivatives of interest as:
$$D_i(x,\boldsymbol{\theta}) \equiv \frac{\partial f_X}{\partial \theta_i} (x|\boldsymbol{\theta}).$$
The characteristic function $\phi$ is defined as:
$$\phi_X(t|\boldsymbol{\theta}) = \int \limits_\mathbb{R} e^{itx} f_X(x|\boldsymbol{\theta}) \ dx,$$
so taking a partial derivative of the characteristic function gives:
$$\begin{align}
\frac{\partial \phi_X}{\partial \theta_i}
(t|\boldsymbol{\theta}) 
&= \frac{\partial}{\partial \theta_i} \int \limits_\mathbb{R} e^{itx} f_X(x|\boldsymbol{\theta}) \ dx \\[6pt]
&= \int \limits_\mathbb{R} e^{itx} \frac{\partial f_X}{\partial \theta_i} (x|\boldsymbol{\theta}) \ dx \\[6pt]
&= \int \limits_\mathbb{R} e^{itx} D_i(x,\boldsymbol{\theta}) \ dx. \\[6pt]
\end{align}$$
This shows that the partial derivatives of the characteristic function are the Fourier transforms of the partial derivatives of the density function.  Consequently, we can obtain the partial derivatives of interest by applying the inverse-Fourier transform to the partial derivatives of the characteristic function:
$$D_i(x,\boldsymbol{\theta}) 
= \mathcal{F}^{-1} \bigg[ \frac{\partial \phi_X}{\partial \theta_i} \bigg]
= \frac{1}{2 \pi} \int \limits_\mathbb{R} e^{-itx} \frac{\partial \phi_X}{\partial \theta_i}
(t|\boldsymbol{\theta}) \ dt.$$
We can write this result in a more succinct fashion for the gradient vector of the density as:
$$\nabla_\boldsymbol{\theta} f_X(x|\boldsymbol{\theta}) 
= \frac{1}{2 \pi} \int \limits_\mathbb{R} e^{-itx} \nabla_\boldsymbol{\theta} \phi_X
(t|\boldsymbol{\theta}) \ dt.$$
Now, in the case you describe, it sounds like you can get the gradient vector for the characteristic function without too much trouble.  You will then need to apply the inverse-Fourier transform to this, which might require you to use numerical methods.  The result will be a function of your parameter $\boldsymbol{\theta}$, so you can then substitute an appropriate estimator for this parameter to get a corresponding estimator for the partial derivatives of interest.
