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