If \(x = f(t)\) and \(y = g(t)\) are differentiable functions of \(t\), so that \(y\) is function of \(x\) and \(\frac{dx}{dt} \ne 0\) then prove that \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).Hence find \(\frac{dy}{dx}\), if \(x = at^2\), \(y = 2at\).

Answer: Proof below; for \(x = at^2\), \(y = 2at\), \(\frac{dy}{dx} = \frac{1}{t}\).

Proof: Let \(\delta x\) and \(\delta y\) be the increments in \(x\) and \(y\) corresponding to a small increment \(\delta t\) in \(t\).

Since \(x = f(t)\) and \(y = g(t)\) are differentiable, they are continuous, so \(\delta x \to 0\) and \(\delta y \to 0\) as \(\delta t \to 0\).

For \(\delta t \ne 0\): \(\frac{\delta y}{\delta x} = \frac{\delta y / \delta t}{\delta x / \delta t}\).

Taking limits as \(\delta t \to 0\) (so \(\delta x \to 0\)):

\(\lim_{\delta x \to 0}\frac{\delta y}{\delta x} = \frac{\lim_{\delta t \to 0}\frac{\delta y}{\delta t}}{\lim_{\delta t \to 0}\frac{\delta x}{\delta t}}\), the denominator being \(\frac{dx}{dt} \ne 0\).

Since \(y\) is a differentiable function of \(x\), \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Hence proved.

Hence part: \(x = at^2 \Rightarrow \frac{dx}{dt} = 2at\); \(y = 2at \Rightarrow \frac{dy}{dt} = 2a\).

\(\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}\) (for \(t \ne 0\)).