If `x=f(t) and y=g(t)`, then `(d^2y)/(dx^2)` is equal to

Options

A. `(f'(t)*g''(t)-g'(t)*f''(t))/([f'(t)]^(3))`
B. `(f'(t)*g''(t)-g'(t)*f''(t))/([f'(t)]^(2))`
C. `(g'(t)*f''(t)-f'(t)*g''(t))/([f'(t)]^(3))`
D. `(g'(t)*f''(t)+f'(t)*g''(t))/([f'(t)]^(3))`

Answer

A

Explanation

Given , `x=f(t) and y=g(t)`
On differentiating both sides w.r.t. 't', we get
`(dx)/(dt)=f'(t) and (dy)/(dt)=g'(t)`
We know that, `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`
`implies(dy)/(dx)=(g'(t))/(f'(t))`
Again, differentiating both sides w.r.t. 'x', we get
`(d^(2)y)/(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))`