>>11650948product rule is d(fg) = d(f) g + f d(g)
so you'd need
d(f) g + f d(g) = d(f) d(g)
for some soultions, let's assume both functions are the same form
2 d(f) f = d(f)^2
2 f = d(f)
in this case, the derivative of the function needs to be equal to itself times 2, meaning f = exp(2x)
d(exp(2x) * exp(2x)) = d(exp(2x)) d(exp(2x))
d(exp(4x)) = 2 exp(2x) 2 exp(2x)
4 exp(4x) = 4 exp(4x)
so at least there is that solution.
