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if x=loge t , t > 0 and y+1=t^2 then (d^...

if `x=log_e t , t > 0` and `y+1=t^2` then `(d^2y)/(dx^2)`

A

`4e^(2x)`

B

`-(1)/(2) e^(-4e)`

C

` - (3)/(4)e^(5x)`

D

`4e^(x)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given , ` x = log _(e) t rArr e^(x) = t `
and ` y + 1 = t^(2) rArr (dx)/(dy) = (1)/(2e^(2x))`
Again , differentiating both sides w.r.t.y, we get
`(d^(2) x)/(dy^(2)) = (1)/(2) e^(-2x) (-2) (dx)/(dy)`
` =- e^(-2x). (1)/(2e^(2x)) = - (1)/(2) e^(-4x)` .
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