Derivative at its Peak.

Now I will post a general solution. But before proceeding the problem we will prove a more general result which is often useful.

We want to express $\dfrac{d^{2}x}{dy^{2}}$ in terms of derivatives of $y$ .

Suppose we have two functions $f(x)$ and $g(x)$ such that $f(x)$ is inverse function of $g(x)$ .

Using the properties of inverse functions:- $g(f(x)) = x$ Differentiating both sides w.r.t $x$ . $g'(f(x))f'(x)=1$ $\implies g'(f(x)) = \dfrac{1}{f'(x)}$ Again differentiating w.r.t. $x$ :- $g''(f(x)) f'(x) = \dfrac{-f''(x)}{(f'(x))^{2}}$ $\implies g''(f(x)) = -\dfrac{f''(x)}{(f'(x))^{3}}$ Above equation can be rewritten in the notation given in above problem as :- $\dfrac{d^{2}x}{dy^{2}} = -\dfrac{\dfrac{d^{2}y}{dx^{2}}}{\bigg( \dfrac{dy}{dx} \bigg) ^{3} }$ The above equation can be rearranged as $\bigg( \dfrac{dy}{dx} \bigg) ^{2} = \Bigg( \dfrac{\dfrac{d^{2}y}{dx^{2}}}{\dfrac{d^{2}x}{dy^{2}}}\Bigg) ^{\dfrac{2}{3}}$ Now we can use this result to manipulate the given expression:- $R^{\dfrac{2}{3}} = \dfrac{1+\bigg(\dfrac{dy}{dx}\bigg)^{2}} {\Bigg( \dfrac{d^{2} y}{dx^{2}}\Bigg) ^{\dfrac{2}{3}}}$ Now we can make use of the above found equation and reach our answer.

Trick is 'R' means Radius of curvature of any path y=f(x) . So for simplicity Take Projectile motion of Particle under gravity (Parabolic) . Now Proceed !

I think this problem is inspired from a problem of IITJEE-2007.