Question

Find `"dy"/"dx"`if, y = `(log "x"^"x") + "x"^(log "x")`

Answer 1

y = `(log "x"^"x") + "x"^(log "x")`

Let u = `(log "x"^"x")` and v = `"x"^(log "x")`

∴ y = u + v

Differentiating both sides w. r. t. x, we get

`"dy"/"dx" = "du"/"dx" + "dv"/"dx"`    ....(i)

Now, u = `(log "x"^"x")`

Taking logarithm of both sides, we get

log u = log `(log "x"^"x")` = x log (log x)

Differentiating both sides w. r. t. x, we get

`"d"/"dx" (log "u") = "x" "d"/"dx" [log (log "x")] + log (log "x") "d"/"dx" ("x")`

∴ `1/"u" * "du"/"dx" = "x"*1/(log "x") * "d"/"dx" (log "x") + log (log "x") * 1`

∴ `1/"u" * "du"/"dx" = "x"*1/(log "x") * 1/"x" + log (log "x")`

∴ `"du"/"dx" = "u" [1/(log "x") + log(log "x")]`

∴ `"du"/"dx" = (log "x"^"x") [1/(log "x") + log(log "x")]`          ....(ii)

v = `"x"^(log "x")`

Taking logarithm of both sides, we get

log v = log `("x"^(log "x"))` = log x (log x)

∴ log v = (log x)² 

Differentiating both sides w.r.t. x, we get

`1/"v" * "dv"/"dx" = 2 log "x" * "d"/"dx" (log "x")`

∴ `1/"v" * "dv"/"dx" = 2 log "x" * 1/"x"`

∴ `"dv"/"dx" = "v"[(2 log "x")/"x"]`

∴ `"dv"/"dx" = "x"^(log "x") [(2 log "x")/"x"]`    ....(iii)

Substituting (ii) and (iii) in (i), we get

`"dy"/"dx" = (log "x"^"x") [1/(log "x") + log(log "x")] + "x"^(log "x") [(2 log "x")/"x"]`