Question

Find `"dy"/"dx"`if, y = (2x + 5)^x 

Answer 1

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)]`