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