Help in proving the identity $\frac d {dt}n_s = - \kappa_st$

I want to prove the identity $\frac d {dt}n_s = - \kappa_st$, where $n_s$ is the signed normal, $t$ is the tangent and $k_s$ is the signed curvature of a unit speed plane curve $\gamma$.

I know that $t \cdot n_s = 0$ so by differentiating I get $\frac d {ds} t \cdot n_s + t \cdot \frac d {ds} n_s = 0$. Now inserting the identity $\frac d {ds} t = \kappa_s n_s$ I get that $\kappa_s + t \cdot \frac d {ds} n_s$ which implies $-\kappa_s = t \cdot \frac d {ds} n_s$.

How can I get that $\frac d {ds}n_s = - \kappa_st$ from this ?

I assume that by $\ n_s$ and $\ t_s$, you mean that everything is parametrised by arc length - so may I first suggest that you set everything to be parametrised by the variable$\ s$ for added clarity in your question (then we will be using operator $\frac{d}{ds}$ instead).

You're close, all you need is one more step. You have,

$\ t.n'=-\kappa$

Then simply note that,

$\ n.n = 1 \implies n'.n=0$

That is, $\ n'$ is parallel to unit tangent$\ t$ (since it is orthogonal to the normal which is orthogonal to the tangent - and we are working in the plane!). So write,

$\ n' := \alpha t$

Plug this into what you already have,

$\alpha=-\kappa$

Giving the desired result,

$\ n'=-\kappa t$