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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Find the marginal demand of a commodity where demand is x and price is y. y = x + 2x2+1 - Mathematics and Statistics

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प्रश्न

Find the marginal demand of a commodity where demand is x and price is y.

y = `("x + 2")/("x"^2 + 1)`

बेरीज
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उत्तर

y = `("x + 2")/("x"^2 + 1)`

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

`"dy"/"dx" = "d"/"dx"(("x + 2")/("x"^2 + 1))`

`= (("x"^2 + 1) * "d"/"dx"("x + 2") - ("x + 2") * "d"/"dx" ("x"^2 + 1))/("x"^2 + 1)^2`

`= (("x"^2 + 1)(1 + 0) - ("x + 2")("2x" + 0))/("x"^2 + 1)^2`

`= (("x"^2 + 1)(1) - ("x + 2")("2x"))/("x"^2 + 1)^2`

`= ("x"^2 + 1 - 2"x"^2 - 4"x")/("x"^2 + 1)^2`

∴ `"dy"/"dx" = (1 - "4x" - "x"^2)/("x"^2 + 1)^2`

Now, by derivative of inverse function, the marginal demand of a commodity is

`"dx"/"dy" = 1/("dy"/"dx")`, where `"dy"/"dx" ne 0`

i.e., `"dx"/"dy" = 1/((1 - 4"x" - "x"^2)/("x"^2 + 1)^2) = ("x"^2 + 1)^2/(1 - 4"x" - "x"^2)`

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Derivatives of Inverse Functions
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पाठ 3: Differentiation - EXERCISE 3.2 [पृष्ठ ९२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 3 Differentiation
EXERCISE 3.2 | Q 2. 2) | पृष्ठ ९२

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