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Question
If xm . yn = (x + y)m+n, prove that `"dy"/"dx" = y/x`
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Solution
Given that: xm . yn = (x + y)m+n
Taking log on both sides
log xm . yn = log (x + y)m+n ......[∵ log xy = log x + log y]
⇒ log xm + log yn = (m + n) log (x + y)
⇒ m log x + n log y = (m + n) log (x + y)
Differentiating both sides w.r.t. x
⇒ `"m" * "d"/"dx" log x + "n" * "d"/"dx" log y = ("m" + "n") "d"/"dx" log (x + y)`
⇒ `"m" * 1/x + "n" * 1/y * "dy"/"dx" = ("m" + "n") * 1/(x + y) (1 + "dy"/"dx")`
⇒ `"m"/x + "n"/y * "dy"/"dx" = ("m" + "n")/(x + y) * (1 + "dy"/"dx")`
⇒ `"m"/x + "n"/y * "dy"/"dx" = ("m" + "n")/(x + y) + ("m" + "n")/(x + y) * "dy"/"dx"`
⇒ `"n"/y * "dy"/"dx" - ("m" + "n")/(x + y) * "dy"/"dx" = ("m" + "n")/(x + y) - "m"/x`
⇒ `("n"/y - ("m" + "n")/(x + y))"dy"/"dx" = ("m" + "n")/(x + y) - "m"/x`
⇒ `(("n"x + "n"y - "m"y - "n"y)/(y(x + y)))"dy"/"dx" = (("m"x + "n"x - "m"x - "m"y)/(x(x + y)))`
⇒ `(("n"x - "m"y)/(y(x + y))) "dy"/"dx" = (("n"x- "m"y)/(x(x + y)))`
⇒ `"dy"/"dx" = ("n"x - "m"y)/(x(x + y)) xx (y(x + y))/("n"x - "m"y)`
⇒ `"dy"/"dx" = y/x`
Hence proved.
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