Question

If $x$ and $y$ are differentiable functions of $t$, then show that
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ if $\frac{dx}{dt}\ne 0$

Answer 1

Let $\delta x$ and $\delta y$ be the increments in $x$ and $y$ respectively, corresponding to increment $\delta t$ in $t$.
Since $x$ and $y$ are differentiable functions of $t$.
$\therefore \underset{\delta t\to 0}{lim}\frac{\delta x}{\delta t}=\frac{dx}{dt}$ ... (i)
and $\underset{\delta t\to 0}{lim}\frac{\delta y}{\delta t}=\frac{dy}{dt}$ ... (ii)
Also, $\delta t\to 0,\delta x\to 0$ ... (iii)
Now $\frac{\delta x}{\delta y}=\frac{\left(\frac{\delta y}{\delta t}\right)}{\left(\frac{\delta x}{\delta t}\right)}(\delta t\ne 0)$
Taking limits as $\delta t\to 0$, we get
$\underset{\delta t\to 0}{lim}\frac{\delta y}{\delta t}=\underset{\delta t\to 0}{lim}\frac{\frac{\delta y}{\delta t}}{\frac{\delta x}{\delta t}}$
$\underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\frac{\underset{\delta t\to 0}{lim}\left(\frac{\delta y}{\delta t}\right)}{\underset{\delta t\to 0}{lim}\left(\frac{\delta x}{\delta t}\right)}$
$\underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$ [form (i) and (ii)]
$\because$ The limits in R.H.S. exist
$\therefore \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=$ exists and is equal to $\frac{dy}{dx}$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$, If $\frac{dx}{dt}\ne 0$.