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Question
Find `"dy"/"dx"`, if y = `2^("x"^"x")`.
Solution
y = `2^("x"^"x")`
Taking logarithm of both sides, we get
`"dy"/"dx" = 2^("x"^"x") * log 2 * "d"/"dx" ("x"^"x")` ...(i)
Let u = xx
log u = log (xx)
∴ log y = 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")`
= `"x" * 1/"x" + log "x" (1)`
∴ `1/"u" * "du"/"dx" = 1 + log "x"`
∴ `"dy"/"dx" = "u"(1 + log "x")`
∴ `"d"/"dx"("x"^"x") = "x"^"x" (1 + log "x")` ...(ii)
Substituting (ii) in (i), we get
`"dy"/"dx" = 2^("x"^"x") * log 2 * "x"^"x" (1 + log "x")`
`"dy"/"dx" = 2^("x"^"x") * "x"^"x" * log 2 (1 + log "x")`
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