let `y=t^(10)+1,` and `x=t^8+1,` then `(d^2y)/(dx^2)` is

Options

A. `(5)/(2)t`
B. `20t^(8)`
C. `(5)/(16t^(6))`
D. none of these

Explanation

`"Here, "y=t^(10)+1 and x=t^(8) +1`
`therefore" "t^(8)=x-1 or t^(2)=(x-1)^(1//4)`
`"So, "y=(x-1)^(5//4)+1`
Differentiating both sides w.r.t. x, we get `(dy)/(dx)=(5)/(4)(x-1)^(1//4)`
Again, differentiating both sides w.r.t. x, we get
`(d^(2)y)/(dx^(2))=(5)/(16)(x-1)^(-3//4)=(5)/(16(x-1)^(3//4))=(5)/(16(t^(2))^(3))=(5)/(16t^(6))`