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Differentiate the following w.r.t.x: (x3-5)5(x3+3)3

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प्रश्न

Differentiate the following w.r.t.x:

`(x^3 - 5)^5/(x^3 + 3)^3`

योग
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उत्तर

Let y = `(x^3 - 5)^5/(x^3 + 3)^3`

Differentiating w.r.t.x, we get

`"dy"/"dx" = "d"/"dx"[(x^3 - 5)^5/(x^3 + 3)^3]`

`"dy"/"dx" = [(x^3 + 3)^3 × "d"/"dx"(x^3 - 5)^5 - (x^3 - 5)^5 "d"/"dx" (x^3 + 3)^3]/[(x^3 + 3)^3]^2`

`"dy"/"dx" = [(x^3 + 3)^3 × 5(x^3 - 5)^4 × "d"/"dx"(x^3 - 5) - (x^3 - 5)^5 × 3(x^3 + 3)^2 × "d"/"dx" (x^3 + 3)]/(x^3 + 3)^6`

`"dy"/"dx" = [(x^3 + 3)^3 × 5(x^3 - 5)^4 × (3x^2 - 0) - (x^3 - 5)^5 × 3(x^3 + 3)^2 × (3x^2 + 0)]/(x^3 + 3)^6`

`"dy"/"dx" = [3x^2(x^3 + 3)^2(x^3 - 5)^4[5(x^3 + 3) - 3(x^3 - 5)]]/(x^3 + 3)^6`

`"dy"/"dx" = [3x^2(x^3 + 3)^2(x^3 - 5)^4[5x^3 + 15 - 3x^3 + 15]]/(x^3 + 3)^6`

`"dy"/"dx" = [3x^2cancel((x^3 + 3)^2)(x^3 - 5)^4(2x^3 + 30)]/(x^3 + 3)^(cancel(6)4)`

`"dy"/"dx" = [3x^2(x^3 - 5)^4(2x^3 + 30)]/(x^3 + 3)^4`

`"dy"/"dx" = [3x^2(x^3 - 5)^4 . 2(x^3 + 15)]/(x^3 + 3)^4`

`"dy"/"dx" = [6x^2(x^3 + 15)(x^3 - 5)^4]/(x^3 + 3)^4`

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अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ १२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 3.04 | पृष्ठ १२

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