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If xm . yn = (x + y)m+n, prove that dydxdydx=yx

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Question

If xm . yn = (x + y)m+n, prove that `"dy"/"dx" = y/x`

Sum
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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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Chapter 5: Continuity And Differentiability - Exercise [Page 113]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 80. (i) | Page 113

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