Curvature Formula Proof By Definition Question: Use Definition 3.2 to prove Theorem 3.4. 
Definition 3.2 “The signed curvature $k(s)$ of a plane curve $ \alpha: I \rightarrow \mathbb{R^2}, \alpha(u)=(x(u),y(u))$ is defined by $t’(s)=k(s)n(s)$ (where $t(s),n(s)$ are the unit tangent and normal vectors, respectively).
Theorem 3.4 “Let $ \alpha: I \rightarrow \mathbb{R^2}, \alpha(u)=(x(u),y(u))$, be a regular curve (not necessarily parametrised by arc length). Then: 
$$k(u)=\frac{x’(u)y’’(u)—x’’(u)y’(u)}{(x’(u)^2+y’(u)^2)^{\frac{3}{2}}}$$
Below is my attempt. I’m not sure why my denominator has the incorrect index. Clearly, it must be small faux pas but I can’t find it. I’ve looked at other proofs here relating to the curvature (as many as I could, there are a lot) but none seem to be exactly the same as this. I apologise in advance if this is a duplicate. 

 A: As I said in my last comment, the formula $\mathbf{t}'(s) = k(s)\mathbf{n}(s)$ is valid only for the arc- length parametrization. The correct proof for the arbitrary parameter is done below.
Consider the plane curve $\mathbf{r}(u) = (x(u), y(u))$, where $u$ is an arbitrary parameter, and let $s$ be the arc-length parameter. Then, by the chain rule and as $\mathbf{t}(s) = \mathbf{r}'(s)$,
$$\mathbf{r}'(u) =\mathbf{r}'(s) \dfrac{ds}{du} = \dfrac{ds}{du} \mathbf{t}(s)$$
$$\implies \mathbf{t}(s) = \mathbf{r}'(u) \dfrac{1}{\frac{ds}{du}}$$
$$\implies \mathbf{t}'(s) = k \ \mathbf{n}(s) = \mathbf{r}''(u)\dfrac{1}{\left(\frac{ds}{du}\right)^2} - \mathbf{r}'(u) \dfrac{\frac{d^2s}{du^2}}{\left(\frac{ds}{du}\right)^3}$$
$$\implies k \ \mathbf{n}(s) = \dfrac{1}{\left(\frac{ds}{du}\right)^3}\left(\mathbf{r}''(u) \dfrac{ds}{du} - \mathbf{r}'(u) \dfrac{d^2s}{du^2}\right)$$
Now, $\dfrac{ds}{du} = \sqrt{x'(u)^2 + y'(u)^2} \implies \dfrac{d^2 s}{du^2} = \dfrac{x'(u)x''(u) + y'(u)y''(u)}{\sqrt{x'(u)^2 + y'(u)^2}} = (x'(u)x''(u) + y'(u)y''(u))\dfrac{1}{\frac{ds}{du}}$
Using these in the above equation gives,
$$\implies k \ \mathbf{n}(s) = \dfrac{1}{\left(\frac{ds}{du}\right)^4}\left(\mathbf{r}''(u) \left(\dfrac{ds}{du}\right)^2 - \mathbf{r}'(u) \dfrac{d^2s}{du^2} \dfrac{ds}{du}\right)$$
$$\implies k \ \mathbf{n}(s) = \dfrac{1}{\left(\frac{ds}{du}\right)^4}\left(x''(u)\left(\dfrac{ds}{du}\right)^2 - x'(u)\dfrac{d^2s}{du^2} \dfrac{ds}{du}, y''(u)\left(\dfrac{ds}{du}\right)^2 - y'(u)\dfrac{d^2s}{du^2} \dfrac{ds}{du}\right)$$
$$\implies k \ \mathbf{n}(s) = \dfrac{1}{\left(\frac{ds}{du}\right)^4}\left(x''(u)y'(u)^2 - x'(u)y'(u)y''(u), \ y''(u)x'(u)^2 - x'(u)y'(u)x''(u)\right)$$
$$\implies k \ \mathbf{n}(s) = \dfrac{1}{\left(\frac{ds}{du}\right)^4}\big(y'(u)(x''(u)y'(u) - x'(u)y''(u)), \ x'(u)(y''(u)x'(u) - y'(u)x''(u))\big)$$
Taking the magnitude of both sides,
$$k = \dfrac{1}{\left(\frac{ds}{du}\right)^4}\sqrt{(y'(u))^2(x''(u)y'(u) - x'(u)y''(u))^2 + (x'(u))^2(y''(u)x'(u) - y'(u)x''(u))^2}$$
$$\implies k = \dfrac{1}{\left(\frac{ds}{du}\right)^4} \cdot \dfrac{ds}{du}(x'(u)y''(u) - y'(u)x''(u))$$
$$\implies k = \dfrac{x'(u)y''(u) - y'(u)x''(u)}{\left(\frac{ds}{du}\right)^3} = \dfrac{x'(u)y''(u) - y'(u)x''(u)}{(x'(u)^2 + y'(u)^2)^{\frac{3}{2}}}$$
