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Question
If y = f(x) is a differentiable function of x, then show that `(d^2x)/(dy^2) = -(dy/dx)^-3.("d^2y)/(dx^2)`.
Solution
If y = f(x) is a differentiable function of x such that inverse function x = f–1(y) exists, then `"dx"/"dy" = (1)/((dy/dx)), "where" "dy"/"dx" ≠ 0`
∴ `(d^2x)/(dy^2) = "d"/"dy"(dx/dy)`
= `"d"/"dy"[(1)/((dy/dx))]`
= `"d"/"Dx"(dy/dx)^-1 xx "dx"/dy"`
= `-1(dy/dx)^-2."d"/"dx"(dy/dx) xx (1)/((dy/dx)`
= `-(dy/dx)^-2.(d^2y)/(dx^2).(dy/dx)^-1`
∴ `(d^2x)/(dy^2) = -(dy/dx)^-3.(d^2y)/(dx^2)`.
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