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Question
If y = tan–1x, find `("d"^2y)/("dx"^2)` in terms of y alone.
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Solution
Given that: y = tan–1x
⇒ x = tan y
Differentiating both sides w.r.t. y
`"dx"/"dy"` = sec2y
⇒ `"dy"/'dx" = 1/(sec^2y)` = cos2y
Again differentiating both sides w.r.t. x
⇒ `"d"/"dx"("dy"/"dx") = "d"/"dx"(cos^2y)`
⇒ `("d"^2y)/("dx"^2) = 2cos y * "d"/"dx" (cos y)`
⇒ `("d"^2y)/("dx"^2) = 2cos y(- siny) * "dy"/"dx"`
⇒ `("d"^2y)/("dx"^2) = - 2sin y cos y * cos^2 y`
∴ `("d"^2y)/("dx"^2)` = – 2 sin y cos3y
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