Question

Find `"dy"/"dx"` if y = `sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))`

Answer 1

y = `sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))`

`= ("3x" - 4)^(3/2)/(("x + 1")^(4/2)*("x + 2")^(1/2))`

Taking logarithm of both sides, we get

log y = log[`("3x" - 4)^(3/2)/(("x + 1")^(4/2)*("x + 2")^(1/2))]`

`= log ("3x" - 4)^(3/2) - [log ("x + 1")^2 + log ("x + 2")^(1/2)]`

`= 3/2 log("3x" - 4) - 2log ("x + 1") - 1/2 log ("x + 2")`

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

`1/"y" * "dy"/"dx" = 3/2 * "d"/"dx" [log (3"x" - 4)] - 2 "d"/"dx" [log ("x + 1")] - 1/2 * "d"/"dx" [log ("x + 2")]`

`= 3/2 * 1/("3x" - 4) * "d"/"dx" ("3x" - 4) - 2 * 1/("x + 1") * "d"/"dx" ("x + 1") - 1/2 * 1/("x + 2") * "d"/"dx" ("x + 2")`

∴ `1/"y" * "dy"/"dx" = 3/(2("3x - 4")) xx 3 - 2/("x + 1") xx 1 - 1/(2 ("x + 2")) xx 1`

∴ `1/"y" * "dy"/"dx" = 9/(2("3x - 4")) - 2/("x + 1") - 1/(2 ("x + 2"))`

∴ `"dy"/"dx" = "y"/2 [9/"3x - 4" - 4/"x + 1" - 1/"x + 2"]`

∴ `"dy"/"dx" = 1/2 sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2"))) [9/"3x - 4" - 4/"x + 1" - 1/"x + 2"]`