1.

If `y=(tan^(-1)x)^2`, then `(x^2+1)^2 (d^2y)/(dx^2) + 2x(x^2+1) dy/dx=`A. 4B. 2C. 1D. 0

Answer» Correct Answer - A
We have,
`y=(tan^(-1)x)^(2)`
On differentiating w.r.t.x, we get
`(dy)/(dx)=(2tan^(-1)x)/(1+x^(2))`
`rArr(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)`
`rArr(1+x^(2))^(2)((dy)/(dx))^(2)=4y" "[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`


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