Note: ${x_p}$ and ${x_{pp}}$ are derivative of x wrt p once and twice, respectively.
I am deriving the curvature for an arbitrary curve C=(x(p),y(p)). The formula for derivation satisfies $\frac{1}{|C_p|}$$\frac{\delta}{\delta p}$($\frac{C_p}{|C_p|}$)=$\kappa$*N.
N=$\frac{(-y_p,x_p)}{\sqrt{x_p^2+y_p^2}}$. is the unit normal to the tangent(tangent to curve).
After doing the derivation according to the formula above, we will have (A,B)=$\kappa$N, for some A and B.
and substituting the x (first) component of N to get $\kappa$, I get $\frac{-x_{pp}*y_p}{(x_p^2+y_p^2)^\frac{3}{2}}$.
Similarly, if I had done same for y component, ie., $\frac{B}{N_y}$, I get
$\frac{-y_{pp}*x_p}{(x_p^2+y_p^2)^\frac{3}{2}}$
The problem is, curvature $\kappa$ is defined as the sum of the two! $\kappa$=$\frac{y_{pp}*x_p-x_{pp}*y_p}{(x_p^2+y_p^2)^\frac{3}{2}}$
Is this because the first and second derivatives are normal to each other? If so, how can I use this fact to show that the summation will still give correct result upon multiplication, somewhere I think $x_{pp}$ and $x_p$ has to multiply and vanish but the derivates being normal to each other only imply that dot product sums should be zero.
I'm going to use more standard notation to make this derivation more generally useful.
Let $\gamma : [a,b] \to \mathbb{R}^2$ be a curve, and assume that $\gamma'(t) \neq 0$. The tangent vector is defined as $$ T = \frac{\gamma'(t)}{\lVert \gamma'(t) \rVert}. $$ $T$ has unit length, so its derivative must be perpendicular to it: $$ T \cdot T = 1 \implies 2T \cdot T' = 0. $$ In higher dimensions, the normal vector $N$ is defined as proportional to $T'$, but in 2D, we don't need to, since we can just specify that $N$ is a rotation of $T$ by $\pi/2$ anticlockwise: i.e. if $T=(a,b)$, $N = (-b,a)$. Then we define the curvature by $$ T' = \kappa N \lVert \gamma'(t) \rVert, $$ where the extra factor of $\lVert \gamma'(t) \rVert$ is needed to make sure that $\kappa$ is independent of rescaling (think about setting $\tau=2t$ and recalculating both sides, for example).
Presumably you know all this already, but now I've fixed the notation and definitions. Now, we calculate the derivatives of $\gamma$ directly to find $\kappa$ in terms of things we know. Write $s'=\lVert\gamma'(t) \rVert$ for compactness. $$\begin{align} \gamma' &= s' T \\ \gamma'' &= s'' T + s' T' = s'' T + s'^2 \kappa N. \end{align} $$ Dotting with $N$, we find $$ \kappa = \frac{\gamma'' \cdot N}{s'^2}. $$ Now we expand this in terms of coordinates: let $\gamma=(x,y)$. Obviously $\gamma'' = (x'',y'')$, and $s'^2 = x'^2+y'^2$. The normal, as noted above, is found from the tangent vector, as $N = (-y',x')/\sqrt{x'^2+y'^2}$. Putting this all together gives $$ \kappa = \frac{(x'',y'')\cdot (-y',x')}{(x'^2+y'^2)^{3/2}} = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}. $$