Question

y = sin^(-1)(x/√1+x²) + cos^(-1)(1/√1+x²) , the value of dy/dx is:.​

Answer 1

Step-by-step explanation:

∵ y = sinֿ¹ [ x.√(1-x) - √x. √(1-x²) ]

∴ y = sinֿ¹ [ (x). √(1-(√x)²) - √x. √(1-(x)²) ] ....... (1)

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Let : (x) = sin α and √x = sin ß. .................... (2)

Then, from (1),

y = sinֿ¹ [ sin α. √(1-sin² ß) - sin ß. √(1-sin² α) ]

. = sinֿ¹ [ sin α. cos ß - sin ß. cos α ]

. = sinֿ¹ [ sin ( α+ß ) ]

. = α + ß

. = sinֿ¹ x + sinֿ¹ √x ............. from (2)

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∴ dy/dx = [ 1 / √(1-x²) ] + [ 1 / √(1-(√x)²) ]· d/dx ( √x )

. . . . . . = [ 1 / √(1-x²) ] + [ 1 / √(1-x) ]· [ 1 / (2√x) ]

. . . . . . .= [ 1 / √(1-x²) ] + { 1 / [ 2√(x-x²) ] }

2 Answers