मराठी

Differentiate xsinx w.r.t. sin x

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प्रश्न

Differentiate `x/sinx` w.r.t. sin x

बेरीज
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उत्तर

Let y = `x/sinx` and z = sin x.

Differentiating both the parametric functions w.r.t. x

`"dy"/"dx" = (sin x * "d"/"dx" (x) - x * "d"/"dx" (sin x))/(sin x)^2`

= `(sin x * 1 - x * cos x)/(sin^2x)`

= `(sinx - x cos x)/(sin^2x)`

`"dz"/"dx"` = cos x

∴ `"dy"/"dz" = ("dy"/"dx")/("dz"/"dx")`

= `((sinx - x cos x)/sin^2x)/cosx`

= `(sinx - xcosx)/(sin^2x cos x)`

= `sinx/(sin^2x cosx) - (xcosx)/(sin^2x cosx)`

= `tanx/(sin^2x) - x/(sin^2x)`

= `(tanx - x)/(sin^2x)`

Hence, `"dy"/"dz" = (tanx - x)/(sin^2x)`.

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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ १११]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 52 | पृष्ठ १११

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