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Sum
Find `"dy"/"dx"` if, y = `root(3)("a"^2 + "x"^2)`
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Solution
y = `root(3)("a"^2 + "x"^2)`
∴ y = `("a"^2 + "x"^2)^(1/3)`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx"[("a"^2 + "x"^2)^(1/3)]`
`= 1/3 ("a"^2 + "x"^2)^(-2/3) * "d"/"dx" ("a"^2 + "x"^2)`
`= 1/3 ("a"^2 + "x"^2)^(-2/3) * (0 + 2"x")`
∴ `"dy"/"dx" = "2x"/3 ("a"^2 + "x"^2)^(-2/3)`
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