Question

If x = f(t) and y = g (t) are differentiable functions of t, then `(d^2y)/(dx^2)` is ______.

Options

• `(f^'(t).g^('')(t) - g^'(t).f^('')(t))/[f^'(t)]^3`

• `(f^'(t).g^('')(t) - g^'(t).f^('')(t))/[f^'(t)]^2`

• `(g^'(t).f^('')(t) - f^'(t).g^('')(t))/[f^'(t)]^3`

• `(g^'(t).f^('')(t) + f^'(t).g^('')(t))/[f^'(t)]^3`

Answer 1

If x = f(t) and y = g (t) are differentiable functions of t, then `(d^2y)/(dx^2)` is `underlinebb((f^'(t).g^('')(t) - g^'(t).f^('')(t))/[f^'(t)]^3`.

Explanation:

We have x = f'(t) and y = g(t)

After differentiating on both sides w.r.t 't', we get

`dx/dt` = f'(t) and `dy/dt` = g'(t)

As,  `dy/dx = (dy/dt)/(dx/dt)`

`\implies dy/dx = g^'(t)/(f^'(t)`

Now, differentiating again both sides w.r.t. 'x', we get

`(d^2y)/(dx^2) = (f^'(t).g^('')(t) - g^'(t).f^('')(t))/(f^'(t))^2. dt/dx`

= `(f^'(t).g^('')(t) - g^'(t).f^('')(t))/(f^'(t))^2. 1/f^'(t)`

= `(f^'(t).g^('')(t) - g^'(t).f^('')(t))/(f^'(t))^3`