Find `"dy"/"dx"`if, y = `"e"^("x"^"x")`
Solution
y = `"e"^("x"^"x")`
Taking the logarithm of both sides, we get
log y = log `"e"^("x"^"x") = "x"^"x" log "e"`
∴ log y = `"x"^"x"`
Differentiating both sides w.r.t.x, we get
`1/"y" * "dy"/"dx" = "d"/"dx" ("x"^"x")` .....(i)
Let u = `"x"^"x"`
Taking logarithm of both sides, we get
log u = `log "x"^"x" = "x" log "x"`
Differentiating both sides w. r. t. x, we get
`1/"u" * "du"/"dx" = "x" * "d"/"dx" (log "x") + log "x" * "d"/"dx"("x")`
∴ `1/"u" * "du"/"dx" = "x" * 1/"x" + log "x" * (1)`
∴ `1/"u" * "du"/"dx"` = 1 + log x
∴ `"du"/"dx" = "u"(1 + log "x")`
∴ `"du"/"dx" = "x"^"x"`(1 + log "x") .....(ii)
Substituting (ii) in (i), we get
`1/"y" * "dy"/"dx" = "x"^"x"`(1 + log x)
∴ `"dy"/"dx" = "y" "x"^"x" (1 + log "x") = "e"^("x"^"x") * "x"^"x" (1 + log "x")`
