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Question
Find the second order derivatives of the following : xx
Solution
y = xx
∴ log y = log xx = x log x
Differentiating both sides w.r.t. x, we get
`(1)/y."dy"/"dx" = "d"/"dx"(xlogx)`
∴ `(1)/y."dy"/"dx" = x."d"/"dx"(logx) + (logx)."d"/"dx"(x)`
= `x/x + (logx) (1)`
= 1 + log x
∴ `"dy"/"dx" = y(1 + logx) = x^x(1 + log x)`
∴ `"d"/"dx"(x^x) = x^x(1 + log x)` ...(1)
∴ `(d^2y)/(dx^2) = "d"/"dx"[x^2(1 + log x)]`
= `x^x."d"/"dx"(1 + log x) + (1 + log x)."d"/"dx"(x^x)`
= `x^x(0 + 1/x) + (1 + logx).x^x(1 + logx)` ...[By (1)]
= xx–1 + xx (1 + log x)2.
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