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Differentiate the following w.r.t. x : x5.tan34xsin23x - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x : `(x^5.tan^3 4x)/(sin^2 3x)`

योग
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उत्तर

Let y = `(x^5.tan^3 4x)/(sin^2 3x)`

Then log y = `log[(x^5.tan^3 4x)/(sin^23x)]`

= logx5 + log tan34x – log sin23x

= 5logx + 3log (tan4x) – 2log (sin3x)

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

`(1)/y."dy"/"dx" = 5"d"/"dx"(logx) + 3"d"/"dx"[log(tan4x)] - 2"d"/"dx"[log(sin3x)]`

= `5 xx (1)/x + 3 xx (1)/(tan4x)."d"/"dx"(tan4x) - 2 xx (1)/(sin3x)."d"/"dx"(sin3x)`

= `5/x + 3 xx (1)/(tan4x) xx sec^2  4x."d"/"dx"(4x) - 2 xx (1)/(sin3x) xx cos3x."d"/"dx"(3x)`

= `5/x + 3.(cos4x)/(sin4x) xx (1)/(cos^2 4x) xx 4 - 2cot3x xx 3`

= `5/x + (24)/(2sin4x.cos4x) - 6cot3x`

∴ `"dy"/"dx" = y[5/x + 24/(sin8x) - 6cot3x]`

= `(x^5.tan^3 4x)/(sin^2 3x)[5/x + 24"cosec"8x - 6cot3x]`.

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Differentiation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
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 1.5 | पृष्ठ ३९

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