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

Options

A. `4e^(2x)`
B. `-(1)/(2)e^(-4x)`
C. `-(3)/(4)e^(5x)`
D. `4e^(x)`

Answer

B

Explanation

Given, `x=log_(e)t`
`impliese^(x)=t and y+1=t^(2)`
`impliesy=e^(2x)-1`
On differentiating w.r.t. y, we get
`2e^(2x)(dx)/(dt)=1`
`implies(dx)/(dy)=(1)/(2e^(2x))`
Again, differentiating 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)`