Question

If x = f(t) and y = g(t) are differentiable functions of t so that y is a differentiable function of x and ${\displaystyle \frac{dx}{dt}}$ ≠ 0 then ${\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{d}}}$.
Hence find ${\displaystyle \frac{dy}{dx}}$ if x = sin t and y = cost

Answer 1

x and y are differentiable functions of t.

Let there be a small increment δt in the value of t.

Correspondingly, there should be small increments δx, δy in the values of x and y, respectively.

As δt → 0, δx → 0, δy → 0

Consider, ${\displaystyle \frac{\delta y}{\delta x}=\frac{\frac{\delta y}{\delta t}}{\frac{\delta x}{\delta t}},\frac{\delta x}{\delta t}}$ ≠ 0
Taking ${\displaystyle \underset{\delta t\to 0}{lim}}$ on both sides, we get

${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\frac{\underset{\delta t\to 0}{lim}\frac{\delta y}{\delta t}}{\underset{\delta t\to 0}{lim}\frac{\delta x}{\delta t}}}$

Since x and y are differentiable functions of t, ${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta t}=\frac{dy}{dt}}$ exists and is finite.

Also, ${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta x}{\delta t}=\frac{dx}{dt}}$ exists and is finite.

∴ ${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)}$

As δt → 0, δx → 0

∴ ${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)}$   .......(i)

Here, R.H.S. of (i) exist and are finite.

Hence, limits on L.H.S. of (i) also should exist and be finite.

∴ ${\displaystyle \underset{\delta t\to 0}{lim}\frac{\delta y}{\delta x}=\frac{dy}{dx}}$ exists and is finite.

∴ ${\displaystyle \frac{dy}{dx}=\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right),\frac{dx}{dt}}$ ≠ 0

Now, x = sin t and y = cos t

∴ ${\displaystyle \frac{dx}{dt}}$ = cos t and ${\displaystyle \frac{dy}{dt}}$ = –sin t

∴ ${\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-\sin t}{\cos t}}$ = – tan t