Question

Find `"dy"/"dx"`if, y = `(1 + 1/"x")^"x"`

Answer 1

y = `(1 + 1/"x")^"x"`

Taking logarithm of both sides, we get

log y = `log(1 + 1/"x")^"x"`

∴ log y = x `log(1 + 1/"x")`

Differentiating both sides w.r.t.x, we get

`1/"y" * "dy"/"dx" = "x" * "d"/"dx" log(1 + 1/"x") + log(1 + 1/"x") * "d"/"dx" ("x")`

∴ `1/"y" * "dy"/"dx" = "x" * 1/(1 + 1/"x") * "d"/"dx" (1 + 1/"x") + log (1 + 1/"x") * (1)`

∴ `1/"y" * "dy"/"dx" = "x"/(("x" + 1)/"x") * (0 - 1/"x"^2) + log (1 + 1/"x")`

∴ `1/"y" * "dy"/"dx" = "x"^2/("x + 1") * ((-1)/"x"^2) + log (1 + 1/"x")`

∴ `1/"y" * "dy"/"dx" = (- 1)/("x + 1") + log (1 + 1/"x")`

∴ `"dy"/"dx" = "y"[(-1)/("x + 1") + log (1 + 1/"x")]`

∴ `"dy"/"dx" = (1 + 1/"x")^"x" * [log (1 + 1/"x") - 1/("x + 1")]`