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Question
Find `"dy"/"dx"`if, y = `"x"^("e"^"x")`
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Solution
y = `"x"^("e"^"x")`
Taking logarithm of both sides, we get
log y = log `"x"^("e"^"x") = "e"^"x" log "x"`
Differentiating both sides w.r.t. x, we get
`1/"y" * "dy"/"dx" = "e"^"x" "d"/"dx" (log "x") + log "x" "d"/"dx" ("e"^"x")`
`= "e"^"x" xx 1/"x" + (log "x")"e"^"x"`
∴ `"dy"/"dx" = "y" * "e"^"x"(1/"x" + log "x") = "x"^("e"^"x") "e"^"x"(1/"x" + log "x")`
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