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Question
If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
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Solution
Let y = (tan–1 x)2 ....(1)
Differentiating (1) w.r.t. x, we get,
`dy/dx = 2 tan^-1 x . 1/(1 + x^2)`
`(d^2y)/dx^2 = 2 [tan^-1 x (({1 + x^2} . 0 - 2x))/(1 + x^2)^2 + 1/ (1 + x^2). 1/ (1 + x^2)]`
= `2 [(-2x tan^-1 x)/ (1 +x^2)^2 + 1/ (1 + x^2)^2]`
= `2 [(-2x tan^-1 x + 1)/ (1 + x^2)^2]`
Now, `(x^2 + 1)^2 (d^2y)/dx^2 + 2x (x^2 + 1) dy/dx`
= `(x^2 + 1)^2 . 2 [(-2x tan^-1x + 1)/ (1 + x^2)^2] + 2x (x^2 + 1). 2tan ^-1 x. 1/ (1 + x^2)`
= −4x tan−1 x + 2 + 4x tan−1 x
= 2
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