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Find dy/dx in the following: y = sin-1(2x/1+x2) - Mathematics

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प्रश्न

Find `bb(dy/dx)` in the following:

`y = sin^(-1)((2x)/(1+x^2))`

योग
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उत्तर

y = `sin^-1 ((2x)/(1 + x^2))`

Let, x = tan θ

⇒ θ = tan−1 x

∴ `y = sin^-1 ((2  tan theta)/(1 + tan^2 theta))`

= `sin^-1 (sin 2 theta)  ... [because sin 2 theta = (2 tan theta)/(1 + tan^2 theta)]`

= 2 θ 

y = 2 tan−1 x

∴ `dy/dx = 2 d/dx tan^-1 x`

`dy/dx = 2/(1 + x^2)`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity and Differentiability - Exercise 5.3 [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.3 | Q 9 | पृष्ठ १६९

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