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Question
If y = `("x" + sqrt("x"^2 - 1))^"m"`, then `("x"^2 - 1) "dy"/"dx"` = ______.
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Solution
If y = `("x" + sqrt("x"^2 - 1))^"m"`, then `("x"^2 - 1) "dy"/"dx"` = my.
Explanation:
y = `("x" + sqrt("x"^2 - 1))^"m"`
Differentiating both sides w.r.t. x, we get
`"dy"/"dx" = "m" ("x" + sqrt("x"^2 - 1))^"m - 1" * "d"/"dx" ("x" + sqrt("x"^2 - 1))`
`= "m" ("x" + sqrt("x"^2 - 1))^"m"/("x" + sqrt("x"^2 - 1))^1 * [1 + 1/(2sqrt("x"^2 - 1)) * "d"/"dx" ("x"^2 - 1)]`
`= "my"/("x" + sqrt("x"^2 - 1)) xx [(1 + 1/(2sqrt("x"^2 - 1))) ("2x")]`
`= "my"/("x" + sqrt("x"^2 - 1)) xx (1 + "x"/sqrt("x"^2 - 1))`
∴ `"dy"/"dx" = "my"/("x" + sqrt("x"^2 - 1)) xx (sqrt("x"^2 - 1) + "x")/sqrt("x"^2 - 1)`
∴ `"dy"/"dx" = "my"/sqrt("x"^2 - 1)`
∴ `sqrt("x"^2 - 1) * "dy"/"dx" = "my"`
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