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Differentiate the following w.r.t. x: tan-1(x1+6x2)+cot-1(1-10x27x) - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x:

`tan^-1(x/(1 + 6x^2)) + cot^-1((1 - 10x^2)/(7x))`

योग
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उत्तर

Let y = `tan^-1(x/(1 + 6x^2)) + cot^-1((1 - 10x^2)/(7x))`

= `tan^-1(x/(1 + 6x^2)) + cot^-1((1 - 10x^2)/(7x))          ...[∵ cot^-1x = tan^-1(1/x)]`

= `tan^-1(x/(1 + 6x^2)) + tan^-1((7x)/(1 - 10x^2))           ...[∵ cot^-1x = tan^-1(1/x)]`

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

= tan–13x – tan–12x + tan–15x + tan–12x
= tan–13x + tan–15x

∴ `"dy"/"dx" = "d"/"dx"[tan^-1 3x + tan^-1 5x]`

= `"d"/"dx"(tan^-1 3x) + "d"/"dx"(tan^-1 5x)`

= `(1)/(1 + (3x)^2)."d"/"dx"(3x) + (1)/(1 + (5x)^2)."d"/"dx"(5x)`

= `(1)/(1 + 9x^2) xx 3 xx 1 + (1)/(1 + 25x^2) xx 5 xx 1`

= `(3)/(1 + 9x^2) + (5)/(1 + 25x^2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Miscellaneous Exercise 1 (II) [पृष्ठ ६४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 4.5 | पृष्ठ ६४

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