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Differentiate the following w.r.t. x : sin-1(1+x22) - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x :

`sin^-1(sqrt((1 + x^2)/2))`

योग
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उत्तर

Let y = `sin^-1(sqrt((1 + x^2)/2))`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[sin^-1(sqrt((1 + x^2)/2))]`

= `(1)/(sqrt(1 - (sqrt((1 + x^2)/2)))^2)."d"/"dx"(sqrt((1 + x^2)/2))`

= `(1)/(sqrt((1 - (1 + x^2)/2))^2) . 1/(2sqrt((1 + x^2)/2)) . 1/2 . 2x`

= `(1)/(((sqrt(2 - 1 + x^2))/sqrt2)^2) . 1/((2sqrt(1 + x^2))/sqrt2) . 1/2 . 2x`

= `(1)/((sqrt(1 - x^2)/sqrt2)^2) . 1/((2sqrt(1 + x^2))/sqrt2) . 1/2 . 2x`

= `(1)/(((1 - x^2)/2)) . 1/(2((sqrt(1 + x^2))/sqrt2)) . 1/2 . 2x`

= `2/((1 - x^2)) . sqrt2/(2sqrt(1 + x^2)) . 1/2 . 2x`

= `(2 . sqrt2 . 2x)/((1 - x^2) . 2 . 2 . sqrt(1 + x^2))`

= `(sqrt2 . x)/((1 - x)^2 (sqrt(1 + x^2))`

= `x/sqrt((1 - x^2)(1 + x^2)`

= `x/sqrt(1 - x^4)`

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Differentiation
  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 6.06 | पृष्ठ २९

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