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प्रश्न
Find `"dy"/"dx"` if y = `"x"^"x" + ("7x" - 1)^"x"`
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उत्तर
y = `"x"^"x" + ("7x" - 1)^"x"`
Let u = xx and v = `("7x" - 1)^"x"`
∴ y = u + v
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "du"/"dx" + "dv"/"dx"` ....(i)
Now, u = xx
Taking logarithm of both sides, we get
log u = log(xx)
∴ log u = 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
∴ `"du"/"dx"` = u(1 + log x)
∴ `"d"/"dx" ("x"^"x") = "x"^"x"`(1 + log x) ....(ii)
Also, v = (7x – 1)x
Taking logarithm of both sides, we get
log v = log(7x - 1)x
∴ log v = x. log(7x – 1)
Differentiating both sides w.r.t.x, we get
`1/"v" * "dv"/"dx" = "x" * "d"/"dx" log ("7x" - 1) + log ("7x" - 1) * "d"/"dx"`(x)
`= "x" * 1/("7x" - 1) * "d"/"dx" (7"x" - 1) + log (7"x" - 1) * (1)`
∴ `1/"v" * "dv"/"dx" = "x"/(7"x" - 1) (7 - 0) + log (7"x" - 1)`
∴ `"dv"/"dx" = "v"["7x"/(7"x" - 1) + log(7"x" - 1)]`
∴`"dv"/"dx" = (7"x" - 1)^"x" ["7x"/(7"x" - 1) + log(7"x" - 1)]` ....(iii)
Substituting (ii) and (iii) in (i), we get
`"dy"/"dx" = "x"^"x" (1 + log "x") + (7"x" - 1)^"x" [log(7"x" - 1) + "7x"/(7"x" - 1)]`
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