Q.If $x = f(t)$ and $y = g(t)$ are differentiable functions of $t$, then prove that $y$ is a differentiable function of $x$ and
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$, where $\dfrac{\mathrm{d}x}{\mathrm{d}t} \neq 0$
Hence find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ if $x = a\cos^2 t$ and $y = a\sin^2 t$.

Use the limit definition with a common parameter increment $\delta t$, then apply to the given parametric curve.

Let $\delta t$ be a small increment in $t$, producing increments $\delta x,\delta y$ in $x=f(t), y=g(t)$. Since $f,g$ are differentiable (hence continuous), $\delta t\to0 \implies \delta x\to0,\ \delta y\to0$.

For $\delta x\ne 0$:

$\dfrac{\delta y}{\delta x} = \dfrac{\delta y/\delta t}{\delta x/\delta t}$

Taking the limit as $\delta t \to 0$:

$\dfrac{dy}{dx} = \lim_{\delta x\to0}\dfrac{\delta y}{\delta x} = \dfrac{\displaystyle\lim_{\delta t\to0}\delta y/\delta t}{\displaystyle\lim_{\delta t\to0}\delta x/\delta t} = \dfrac{dy/dt}{dx/dt}, \qquad \dfrac{dx}{dt} e0$

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