Question

Find the derivatives of the following:
y = `x^(logx) + (logx)^x`

Answer 1

y = `x^(logx) + (logx)^x`

Let u = x^{log x}, v = (log x)^x

log u = log x^{log x}

log u = (log x)(log x)

log u = (log x)² 

`1/u * ("d"u)/("d"x) = 2 log x xx 1/x`

`("d"u)/("d"x) = 2u (logx)/x`  ........(1)

v = (log x)^x

log v = log (log x)^x

log v = x log (log x)

`1/"v" * ("dv")/("d"x) = x xx 1/logx xx 1/x + log(log x) xx 1`

`("dv")/("d"x) = "v"[1/logx + log(logx)]` ........(2)

y = u + v

`("d"y)/("d"x) = ("d"u)/("d"x) + ("dv")/("d"x)`

`("d"y)/("d"x) = 2u logx/x + "v"[1/logx + log (log x)]`

By equation (1) and (2)

`("d"y)/("d"x) = 2 x^(logx) * logx/x + (logx)^x [1/logx + log(logx)]`