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Question
Find `"dy"/"dx"`if, y = (2x + 5)x
Solution
y = (2x + 5)x
Taking logarithm of both sides, we get
log y = log (2x + 5)x
∴ log y = x * log (2x + 5)
Differentiating both sides w.r.t.x, we get
`1/"y" "dy"/"dx" = "x" * "d"/"dx"[log (2"x" + 5)] + log ("2x" + 5) * "d"/"dx" ("x")`
`= "x" * 1/("2x" + 5) * "d"/"dx" ("2x" + 5) + log (2"x" + 5) * (1)`
`= "x"/("2x" + 5) * (2 + 0) + log (2"x" + 5)`
∴ `1/"y" "dy"/"dx" = "2x"/("2x" + 5) + log ("2x" + 5)`
∴ `"dy"/"dx" = "y"["2x"/("2x" + 5) + log ("2x" + 5)]`
∴ `"dy"/"dx" = ("2x" + 5)^"x" [log ("2x" + 5) + "2x"/("2x" + 5)]`
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