Question

If x = f(t) and y = g(t) are differentiable functions of t, so that y is a function of x and `dx/dt ≠ 0`, then prove that `dy/dx = ((dy/dt))/((dx/dt))`. Hence find `dy/dx`, if y = at² and x = 2at.

Answer 1

Given: x = f(t) and y = g(t).

Let δx and δy be the increments in x and y respectively, corresponding to the increment δt in t. Since x and y are differentiable functions of t,

`dx/dt = lim_(δt -> 0) (δx)/(δt)` and `dy/dt = lim_(δt -> 0) (δy)/(δt)`   ...(1)

Also, as δt → 0, δx → 0   ...(2)

Now, `(δy)/(δx) = ((δy//δt))/((δx//δt))`   ...[δt ≠ 0]

Taking limits as δt → 0, we get

`lim_(δt -> 0) (δy)/(δx) = lim_(δt -> 0) ((δy//δt))/((δx//δt))`

`lim_(δx -> 0) (δy)/(δx) = (lim_(δt -> 0) (δy//δt))/(lim_(δt -> 0) (δx//δt))`   ...[By (1) and (2)]

`lim_(δx -> 0) (δy)/(δx) = ((dy//dt))/((dy//dt))`

∵ The limits in RHS exist,

∴ `lim_(δx -> 0) (δy)/(δx)` exists and is equal to `dy/dx`

∴ `dy/dx = (dy//dt)/(dx//dt)`, if `dx/dt ≠ 0`.

To find `dy/dx`, if y = at² and x = 2at:

y = at², x = 2at

Differentiating x and y w.r.t. t, we get

`dy/dt = d/dt (at^2) = a d/dt (t^2)`

= a × 2t

= 2at

And `dx/dt = d/dt (2at)`

= `2a d/dt (t)`

= 2a × 1

= 2a

∴ `dy/dx = ((dy//dt))/((dx//dt))`

= `(2at)/(2a)`

= t