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Differentiate the following w.r.t.x. : y = x2sinxx+cosx - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t.x. :

y = `(x^2 sin x)/(x + cos x)`

योग
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उत्तर

Let y = `(x^2 sin x)/(x + cos x)`

∴ `("d"y)/("d"x) = "d"/("d"x) [(x^2 sin x)/(x + cos x)]`

= `((x + cos x) "d"/("d"x) (x^2 sin x) - (x^2 sin x) "d"/("dx") (x + cos x))/(x + cos x)^2`

= `((x + cos x) [x^2 "d"/("d"x) (sin x) + (sin x) "d"/("d"x) (x^2)] - (x^2 sin x) (1 - sin x))/(x + cos x)^2`

= `((x + cos x) [x^2 (cos x) + (sin x)(2x)] - x^2 sin x + x^2 sin^2x)/(x + cos x)^2`

= `(x^3 cos x + 2x^2 sin x + x^2 cos^2x + 2x sin x cos x + x^2 sin^2x - x^2 sinx)/(x + cos x)^2`

= `(x^3 cos x + x^2 sin x + x^2 cos^2x + x^2 sin^2x + x sin 2x)/(x + cos x)^2`

= `(x^2[x cos x + sin x + (cos^2x + sin^2x)] + x sin 2x)/(x + cos x)^2`

= `(x^2[x cos x + sin x + 1] + x sin 2x)/(x + cos x)^2`

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अध्याय 9: Differentiation - Exercise 9.2 [पृष्ठ १९२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 9 Differentiation
Exercise 9.2 | Q IV. (5) | पृष्ठ १९२
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