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