Derivative of ratio of convolutions

61 Views Asked by At

Let $(f\ *_P\ w)(x) = \frac{(f^{P+1}\ *\ w)(x)}{(f^{P}\ *\ w)(x)}\\$.

Please help prove / disprove:

$\frac{\partial(f\ *_P\ w)(x)}{\partial f} = \frac{f(x)}{(f^P\ *\ w)(x)} - \frac{f(x)\ .\ (f^{P+1}\ *\ w)(x)}{(f^P\ *\ w)(x)}$

where: '$*$' denotes convolution, '$.$' element-wise multiplication, $f^P$ is the image where each pixel value of $f$ is raised to power $P$.