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Let x=f(t) and y=g(t), where x and y are twice differentiable function. If f'(0)= g'(0) =f''(0) = 2. g''(0) = 6, then the value of `((d^(2)y)/(dx^(2)))_(t=0)` is equal to

A

0

B

1

C

2

D

3

Text Solution

Verified by Experts

`"We have "(dy)/(dx)=(g'(t))/(f'(t))`
`rArr" "(d)/(dx)((dy)/(dx))=(d)/(dx)((g'(t))/(f'(t)))`
`=(d)/(dt)((g'(t))/(f'(t))).(dt)/(dx)`
`=(f'(t)g''(t)-g'(t)f''(t))/((f'(t))^(2))cdot(dt)/(dx)`
`rArr" "((d^(2)y)/(dx^(2)))_(t=0)=(f'(0)g''(0)-g'(0)f''(0))/((f'(0))^(3))`
`=(2xx6-2xx2)/(2^(3))`
`=1`
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