Question

If y = `sin^-1 {xsqrt(1 - x) - sqrt(x) sqrt(1 - x^2)}` and 0 < x < 1, then find `("d"y)/(dx)`

Answer 1

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))`.