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Question
If `"y" ="x"^"x" , "find" "dy"/"dx"`.
Solution
`"y" ="x"^"x"` ......(1)
Taking log on both sides,
log y = log Xx
logy =x log x
Differentiate both sides w.r.t x
`1/"y" . "dy"/"dx" = ("x" xx 1)/"x" + log "x" xx 1`
`"dy"/"dx" = "y"[1 + log "x"]`
=Xx [ 1+ log x] from (1)
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