# If x7⋅y9=(x + y)16, then show that dydx=yx - Mathematics and Statistics

Sum

If "x"^7*"y"^9 = ("x + y")^16, then show that "dy"/"dx" = "y"/"x"

#### Solution

"x"^7*"y"^9 = ("x + y")^16

Taking logarithm of both sides, we get

log "x"^7*"y"^9 = log ("x + y")^16

∴ log "x"^7 + log "y"^9 = 16 log ("x + y")

∴ 7 log x + 9 log y = 16 log (x + y)

Differentiating both sides w.r.t. x, we get

7(1/"x") + 9(1/"y") "dy"/"dx" = 16(1/("x + y")) "d"/"dx" ("x + y")

∴ 7/"x" + 9/"y" "dy"/"dx" = 16/("x + y") (1 + "dy"/"dx")

∴ 7/"x" + 9/"y" "dy"/"dx" = 16/("x + y") + 16/("x + y") "dy"/"dx"

∴ 9/"y" "dy"/"dx" - 16/("x + y") "dy"/"dx" = 16/("x + y") - 7/"x"

∴ (9/"y" - 16/("x + y")) "dy"/"dx" = 16/("x + y") - 7/"x"

∴ [("9x" + "9y" - 16"y")/("y"("x + y"))] "dy"/"dx" = (16"x" - 7"x" - 7"y")/("x"("x + y"))

∴ [("9x" - 7"y")/("y"("x + y"))] "dy"/"dx" = ("9x" - 7"y")/("x"("x + y"))

∴ "dy"/"dx" = ("9x" - 7"y")/("x"("x + y")) xx ("y"("x + y"))/("9x" - 7"y")

∴ "dy"/"dx" = "y"/"x"

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Chapter 3: Differentiation - Miscellaneous Exercise 3 [Page 100]

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Differentiation
Miscellaneous Exercise 3 | Q 4.13 | Page 100

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