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प्रश्न
If y = `sin^-1 {xsqrt(1 - x) - sqrt(x) sqrt(1 - x^2)}` and 0 < x < 1, then find `("d"y)/(dx)`
उत्तर
We have y = `sin^-1 {xsqrt(1 - x) - sqrt(x) sqrt(1 - x^2)}`
Where 0 < x < 1
Put x = sinA and `sqrt(x)` = sinB
Therefore, y = `sin^-1{sin"A" sqrt(1 - sin^2"B") - sin"B"sqrt(1 - sin^2"A")}`
= `sin^-1 {sin "A" cos "B" - sin "B" cos "A"}`
= `sin^-1 {sin("A" - "B")}`
= A – B
Thus y = `sin^-1x - sin^1 sqrt(x)`
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = 1/sqrt(1 - x^2) - 1/sqrt(1 - sqrt((x)^2)) * "d"/("d"x) (sqrt(x))`
= `1/sqrt(1 - x^2) - 1/(2sqrt(x) sqrt(1 - x))`.
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