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

Answer: Proof below; \(\frac{dy}{dx} = -\tan t\ \left(= -\frac{x}{y}\right)\).

Let \(\delta x\) and \(\delta y\) be the increments in \(x\) and \(y\) corresponding to a small increment \(\delta t\) in \(t\). Since \(\delta t \ne 0\),

\(\frac{\delta y}{\delta x} = \frac{\delta y/\delta t}{\delta x/\delta t}\).

As \(x\) and \(y\) are differentiable (hence continuous) functions of \(t\), \(\delta x \to 0\) and \(\delta y \to 0\) as \(\delta t \to 0\). Taking limits as \(\delta t \to 0\):

\(\lim_{\delta x \to 0}\frac{\delta y}{\delta x} = \frac{\lim_{\delta t \to 0}\frac{\delta y}{\delta t}}{\lim_{\delta t \to 0}\frac{\delta x}{\delta t}}\), which is valid because \(\frac{dx}{dt} \ne 0\).

Hence \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Application: \(x = \sin t \Rightarrow \frac{dx}{dt} = \cos t\); \(y = \cos t \Rightarrow \frac{dy}{dt} = -\sin t\).

\(\frac{dy}{dx} = \frac{-\sin t}{\cos t} = -\tan t\) (for \(\cos t \ne 0\)), which can also be written as \(-\frac{x}{y}\).