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प्रश्न
Find `"dy"/"dx"`if, y = `(log "x"^"x") + "x"^(log "x")`
उत्तर
y = `(log "x"^"x") + "x"^(log "x")`
Let u = `(log "x"^"x")` and v = `"x"^(log "x")`
∴ y = u + v
Differentiating both sides w. r. t. x, we get
`"dy"/"dx" = "du"/"dx" + "dv"/"dx"` ....(i)
Now, u = `(log "x"^"x")`
Taking logarithm of both sides, we get
log u = log `(log "x"^"x")` = x log (log x)
Differentiating both sides w. r. t. x, we get
`"d"/"dx" (log "u") = "x" "d"/"dx" [log (log "x")] + log (log "x") "d"/"dx" ("x")`
∴ `1/"u" * "du"/"dx" = "x"*1/(log "x") * "d"/"dx" (log "x") + log (log "x") * 1`
∴ `1/"u" * "du"/"dx" = "x"*1/(log "x") * 1/"x" + log (log "x")`
∴ `"du"/"dx" = "u" [1/(log "x") + log(log "x")]`
∴ `"du"/"dx" = (log "x"^"x") [1/(log "x") + log(log "x")]` ....(ii)
v = `"x"^(log "x")`
Taking logarithm of both sides, we get
log v = log `("x"^(log "x"))` = log x (log x)
∴ log v = (log x)2
Differentiating both sides w.r.t. x, we get
`1/"v" * "dv"/"dx" = 2 log "x" * "d"/"dx" (log "x")`
∴ `1/"v" * "dv"/"dx" = 2 log "x" * 1/"x"`
∴ `"dv"/"dx" = "v"[(2 log "x")/"x"]`
∴ `"dv"/"dx" = "x"^(log "x") [(2 log "x")/"x"]` ....(iii)
Substituting (ii) and (iii) in (i), we get
`"dy"/"dx" = (log "x"^"x") [1/(log "x") + log(log "x")] + "x"^(log "x") [(2 log "x")/"x"]`
Notes
The answer in the textbook is incorrect.
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