Proof of Curvature Formula

The curvature formula for a curve  
\[y=f(x)\]
  is  
\[\frac{d \theta}{ds} = \kappa = \frac{\frac{d^2y}{dx^2}}{(1+ \frac{dy}{dx})^2)^\frac{3}{2}}\]
, where  
\[\theta\]
  is the angle subtended by an arc of length  
\[s\]
. To prove it note that  
\[tan \theta \frac{dy}{dx} \rightarrow \frac{d}{dx}= \frac{d^2y}{dx^2} \rightarrow sec^2 \theta \frac{d \theta}{dx}= \frac{d^2y}{dx^2} \]

But  
\[sec^2 \theta = 1+tan^2 \theta = 1+ (\frac{dy}{dx})^2\]
  so  
\[ (1+ (\frac{dy}{dx})^2) \frac{d \theta}{dx}= \frac{d^2y}{dx^2} \]
  (1)
Now use  
\[\frac{d \theta}{dx} =\frac{d \theta}{ds} \frac{ds}{dx}\]
  and note that  
\[ds= ((dx)^2+(dy)^2)^{\frac{1}{2}} \rightarrow \frac{ds}{dx} =(1+ (\frac{dy}{dx}))^{\frac{1}{2}}\]
  so  
\[\frac{d \theta}{dx}= (1+ (\frac{dy}{dx}))^{\frac{1}{2}}\frac{d \theta }{ds}\]
.
Then (1) becomes  
\[ (1+ (\frac{dy}{dx})^2)^{\frac{3}{2}} \frac{d \theta}{ds}= \frac{d^2y}{dx^2} \rightarrow \frac{d \theta}{ds} = \frac{\frac{d^2y}{dx^2}}{(1+ \frac{dy}{dx})^2)^\frac{3}{2}} \]