If `y=(tan^(-1)x)^2`, then `(x^2+1)^2 (d^2y)/(dx^2) + 2x(x^2+1) dy/dx=`

Options

A. 4
B. 2
C. 1
D. 0

Answer

A

Explanation

We have,
`y=(tan^(-1)x)^(2)`
On differentiating w.r.t. x, we get
`(dy)/(dx)=(2tan^(-1)x)/(1+x^(2))`
`implies(1+x^(2))(dy)/(dx)=2tan^(-1)x`
On squaring both sides, we get
`(1+x^(2))^(2)((dy)/(dx))^(2)+4(tan^(-1)x)^(2)`
`implies(1+x^(2))^(2)((dy)/(dx))^(2)=4" "[becausey=(tan^(-1)x)^(2)]`
Again, differentiating w.r.t.x, we get
`(1+x^(2))^(2)(2(dy)/(dx)*(d^(2)y)/(dx^(2)))+2(1+x^(2))(2x)((dy)/(dx))^(2)`
`=4(dy)/(dx)`
On dividing both sides by `2(dy)/(dx)`, we get
`(1+x^(2))^(2)((d^(2)y)/(dx^(2)))+2x(1+x^(2))(dy)/(dx)=4`