If `x=logt and y=t^(2)-1`, then what is `(d^(2)y)/(dx^(2))` at t = 1 equal to?A. 2B. 3C. `-4`D. 4

If `x=logt and y=t^(2)-1`, then what is `(d^(2)y)/(dx^(2))` at t = 1 e

1.	If `x=logt and y=t^(2)-1`, then what is `(d^(2)y)/(dx^(2))` at t = 1 equal to?A. 2B. 3C. `-4`D. 4
Answer» Correct Answer - D Let `x=log t and y=t^(2)-1` `x=logt` `rArr 2x=2log t` `rArr 2x=log t^(2)` `rArr 2x=log (y+1)rArr e^(2x)=y+1` On differentiating w.r.t. x, twice, we get `e^(2x)2=(dy)/(dx) rArr 4e^(2x)=(d^(2)y)/(dx^(2))` At t = 1, x = 0 `(d^(2)y)/(dx^(2))=4e^(2(0))=4" "(because e^(0)=1)`