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Differentiate the following w.r.t. x : 4x-1(2x+3)(5-2x)23 - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x : `root(3)((4x - 1)/((2x + 3)(5 - 2x)^2)`

योग
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उत्तर

Let y = `root(3)((4x - 1)/((2x + 3)(5 - 2x)^2)`

Then log y = `log[(4x - 1)/((2x + 3)(5 - 2x)^2)]^(1/3)`

= `(1)/(3)log[(4x - 1)/((2x + 3)(5 - 2x)^2)]`

= `(1)/(3)[log(4x - 1) - log(2x + 3)(5 - 2x)^2]`

= `(1)/(3)log(4x - 1) - (1)/(3)log(2x + 3) - (2)/(3)log(5 - 2x)`
Differentiating both sides w.r.t. x, we get
`(1)/y."dy"/"dx" = (1)/(3)"d"/"dx"[log(4x - 1)] - (1)/(3)"d"/"dx"[log(2x + 3)] - (2)/(3)"d"/"dx"[log(5 - 2x)]`

= `(1)/(3) xx (1)/(4x - 1)."d"/"dx"(4x - 1) - (1)/(3) xx (1)/(2x + 3)."d"/"dx"(2x + 3) - (2)/(3) xx (1)/(5 - 2x)."d"/"dx"(5 - 2x)`

= `(1)/(3(4x - 1)). (4 xx 1 - 0) - (1)/(3(2x + 3)).(2 xx 1 + 0) - (2)/(3(5 - 2x)).(0 - 2 xx 1)`

∴ `"dy"/"dx" = y[(4)/(3(4x - 1)) - (2)/(3(2x + 3)) + (4)/(3(5 - 2x))]`

= `root(3)((4x - 1)/((2x + 3)(5 - 2x)^2))[(4)/(3(4x - 1)) - (2)/(3(2x + 3)) + (4)/(3(5 - 2x))]`.

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अध्याय 1: Differentiation - Exercise 1.3 [पृष्ठ ३९]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.3 | Q 1.2 | पृष्ठ ३९

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