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Differentiate the following w.r.t. x : tan-1[1+cos(x3)sin(x3)]

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प्रश्न

Differentiate the following w.r.t. x : `tan^-1[(1 + cos(x/3))/(sin(x/3))]`

योग
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उत्तर

Let y = `tan^-1[(1 + cos(x/3))/(sin(x/3))]`

= `tan^-1[(2cos^2(x/6))/(2sin(x/6)cos(x/6))]`

= `tan^-1[cot(x/6)]`

= `tan^-1[tan(pi/2 - x/6)]`

= `pi/(2) - x/(6)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(pi/2 - x/6)`

= `"d"/"dx"(pi/2) - (1)/(6)"d"/"dx"(x)`

= `0 - (1)/(6) xx 1`

= `-(1)/(6)`.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.2 [पृष्ठ २९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 7.07 | पृष्ठ २९

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