I have the following equation:
$$r_t=kr_s^{3}+r_s$$
I need to get $r_s'$, i.e. derivative of $r_s$ w.r.t. $k$, provided that value of $r_t'$ is already known. Taking derivatives of both sides w.r.t. to $k$ leads to:
$$r_t'=r_s^3+3kr_s^2\cdot r_s'+r_s'$$
by rearranging, I get:
$$r_s'=\frac{r_t'-r_s^3}{3kr_s^2+1}$$
This formula works as long as denominator is not equal to zero. But how to compute $r_s'$ when the denominator $3kr^2+1=0$ ?
I found a case where both $r_t$ and corresponding $r_s$ exist (found numerically), but the above formula for derivative fails due to division by zero.
One obvious option is tha $r_s'$ fails to exist at those points. On the other hand, if we know that $r_s'$ exists, this can only happen if also the numerator is $0$, i.e., $$r_s^3=r_t'.$$ The left hand side of this is differentiable and we conclude that $r_t''$ exists at this point and $$ \tag13r_s'r_s^2=r_t''$$ i.e., $$r_s'=\frac{r_t''}{3r_s^2}\qquad \text{if }r_s'\text{ exists and }3kr_s^2+1=0.$$ (This cannot fail again as $3r_s^2=0$ implies $3kr_s^2+1\ne 0$.)