Question

Find `"dy"/"dx"`if, y = `root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2))`

Answer 1

y = `root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2))`

`= ("3x" - 1)^(1/3)/(("2x" + 3)^(1/3)*(5 - "x")^(2/3))`

Taking logarithm of both sides, we get

log y = `log[("3x" - 1)^(1/3)/(("2x" + 3)^(1/3)*(5 - "x")^(2/3))]`

`= log ("3x" - 1)^(1/3) - [log ("2x" + 3)^(1/3) + log (5 - "x")^(2/3)]`

`= 1/3 log ("3x" - 1) - [1/3 log ("2x" + 3) + 2/3 log (5 - "x")]`

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

`1/"y" "dy"/"dx" = 1/3 * "d"/"dx" [log ("3x" - 1)] - 1/3 * "d"/"dx" [log (2"x" + 3)] - 2/3 * "d"/"dx" [log (5 - "x")]`

`= 1/3 * 1/("3x" - 1)*"d"/"dx" ("3x" - 1) - 1/3 * 1/("2x + 3") * "d"/"dx" ("2x" + 3) - 2/3 * 1/("5 - x") * "d"/"dx" (5 - "x")`

`= 1/(3(3"x" - 1)) xx 3 - 1/(3(2"x" + 3)) xx 2 - 2/(3(5 - "x")) xx -1`

∴ `1/"y" "dy"/"dx" = 1/("3x" - 1) - 2/(3("2x" + 3)) + 2/(3(5 - "x"))`

∴ `"dy"/"dx" = "y"/3 [3/("3x" - 1) - 2/("2x" + 3) + 2/(5 - "x")]`

∴ `"dy"/"dx" = 1/3 * root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2)) [3/("3x" - 1) - 2/("2x" + 3) + 2/(5 - "x")]`