Question

Find the second order derivatives of the following : x^x 

Answer 1

y = x^x
∴ log y = log x^x = 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)]
= x^x–1 + x^x (1 + log x)².